GSEB Class 11 Statistics Solutions Chapter 6 Permutations, Combinations and Binomial Expansion Exercise 6.2

Question 1. Obtain the values of the following:
(1) \( ^{10}P_2 \)
(2) \( ^{50}P_2 \)
(3) \( ^8P_7 \)
(4) \( ^9P_9 \)
Answer:
(1) Let's obtain the value of \( ^{10}P_2 \):
\( ^{10}P_2 = \frac{10!}{(10-3)!} \)
\( = \frac{10!}{7!} \)
\( = \frac{10 \times 9 \times 8 \times 7!}{7!} \)
\( = 10 \times 9 \times 8 = 720 \)
\( \therefore ^{10}P_2 = 720 \)

Alternative method:
\( ^{10}P_2 = 10(10-1)(10-2) \)
\( = 10 \times 9 \times 8 = 720 \)
\( \therefore ^{10}P_2 = 720 \)

(2) Let's obtain the value of \( ^{50}P_2 \):
\( ^{50}P_2 = \frac{50!}{(50-2)!} \)
\( = \frac{50!}{48!} \)
\( = \frac{50 \times 49 \times 48!}{48!} = 50 \times 49 = 2450 \)
\( \therefore ^{50}P_2 = 2450 \)

Alternative method:
\( ^{50}P_2 = 50(50-1) \)
\( = 50 \times 49 = 2450 \)
\( \therefore ^{50}P_2 = 2450 \)

(3) Let's obtain the value of \( ^8P_7 \):
\( ^8P_7 = \frac{8!}{(8-7)!} \)
\( = \frac{8!}{1!} \)
\( = 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 40320 \)
\( \therefore ^8P_7 = 40320 \)

Alternative method:
\( ^8P_7 = 8(8-1)(8-2)(8-3)(8-4)(8-5)(8-6) \)
\( = 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 = 40320 \)
\( \therefore ^8P_7 = 40320 \)

(4) Let's obtain the value of \( ^9P_9 \):
\( ^9P_9 = \frac{9!}{(9-9)!} \)
\( = \frac{9!}{0!} \)
\( = 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 362880 \)
\( \therefore ^9P_9 = 362880 \)
In simple words: Permutations involve arranging items where the order matters. For \( ^nP_r \), you multiply 'r' decreasing numbers starting from 'n'. If 'r' is equal to 'n', it is simply 'n' factorial. Always remember that \( 0! \) equals 1.

Exam Tip: Always make sure to use the correct formula for permutations and factorials. For \( ^nP_r \), the formula is \( \frac{n!}{(n-r)!} \).

Question 2. If \( ^nP_3 = 990 \), then find the value of n.
Answer:
We are given the equation:
\( ^nP_3 = 990 \)
We know that \( ^nP_3 = n(n-1)(n-2) \). So, we can write:
\( \therefore n(n-1)(n-2) = 990 \)
To find 'n', we need to express 990 as a product of three consecutive integers. We observe that \( 11 \times 10 \times 9 = 990 \).
\( \therefore n(n-1)(n-2) = 11 \times 10 \times 9 \)
Comparing this with the permutation formula, we get:
\( \therefore n(n-1)(n-2) = 11(11-1)(11-2) \)
Thus, by comparing the terms, the value of n is 11.
In simple words: To find 'n', look for three numbers that multiply to 990. Once you find them, the biggest number is 'n'.

Exam Tip: For these types of problems, try to factorize the given number into a product of consecutive integers to easily find the value of n.

Question 3. If \( ^9P_r = 3024 \), find the value of r.
Answer:
The general formula for permutation is:
\( ^nP_r = \frac{n!}{(n-r)!} \)
Given \( n=9 \) and \( ^9P_r = 3024 \), we substitute these values into the formula:
\( ^9P_r = \frac{9!}{(9-r)!} \)
\( \therefore 3024 = \frac{362880}{(9-r)!} \)
Now, we need to solve for \( (9-r)! \):
\( \therefore (9-r)! = \frac{362880}{3024} \)
\( \therefore (9-r)! = 120 \)
We know that \( 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \). So, we can write:
\( \therefore (9-r)! = 5! \)
Comparing both sides, we get:
\( \therefore 9-r = 5 \)
Now, we solve for r:
\( r = 9-5 \)
\( r = 4 \)
Hence, the value of r is 4.
In simple words: We're finding 'r' when we know how many ways 9 things can be arranged 'r' at a time. We use the permutation formula and match the result to a factorial to solve for 'r'.

Exam Tip: Be familiar with common factorial values (e.g., \( 3!=6, 4!=24, 5!=120 \)) as they can greatly speed up solving these problems.

Question 4. If \( 3 \cdot ^{n+3}P_4 = 5 \cdot ^{n+2}P_4 \), then find the value of n.
Answer:
We are given the equation:
\( 3 \cdot ^{n+3}P_4 = 5 \cdot ^{n+2}P_4 \)
Using the definition of \( ^nP_r = n(n-1)...(n-r+1) \), we expand both sides:
\( \implies 3(n+3)(n+3-1)(n+3-2)(n+3-3) = 5(n+2)(n+2-1)(n+2-2)(n+2-3) \)
\( \implies 3(n+3)(n+2)(n+1)n = 5(n+2)(n+1)n(n-1) \)
We can cancel the common terms \( (n+2)(n+1)n \) from both sides:
\( \implies 3(n+3) = 5(n-1) \)
Now, we expand and solve the linear equation:
\( \implies 3n+9 = 5n-5 \)
\( \implies 9+5 = 5n-3n \)
\( \implies 14 = 2n \)
\( \implies n = \frac{14}{2} \)
\( \implies n = 7 \)
The value of n is 7.
In simple words: We expand both permutation terms, simplify the equation by canceling common parts, and then solve for 'n' using basic algebra.

Exam Tip: When dealing with permutation equations, expand the terms carefully and simplify by canceling common factors before solving. Be mindful of algebraic rules.

Question 5. In how many ways can 4 persons be arranged in a row?
Answer:
We need to arrange 4 persons in a row. This is a permutation problem where the number of items (n) is 4 and the number of items to arrange at a time (r) is also 4.
\( \therefore n = 4, r = 4 \)
The total number of permutations is given by \( ^4P_4 \).
\( \therefore \text{Total permutations} = ^4P_4 \)
We know that when \( n=r \), \( ^nP_n = n! \). So,
\( = 4! \)
\( = 4 \times 3 \times 2 \times 1 \)
\( = 24 \)
Thus, 4 persons can be arranged in a row in 24 different ways.
In simple words: To arrange 4 people in a line, you multiply the numbers from 4 down to 1. This gives you all the possible different orders.

Exam Tip: Remember that arranging 'n' items in a row is simply calculating 'n' factorial (n!).

Question 6. How many six-digit numbers can be formed using all the digits 1, 2, 3, 0, 7, 9?
Answer:
We want to form six-digit numbers using all the digits 1, 2, 3, 0, 7, 9. For a number to be a six-digit number, the first digit cannot be 0.
\( \therefore \) Excluding digit 0, one of the five digits (1, 2, 3, 7, 9) can be placed at the first place in \( ^5P_1 \) ways.
Now, the remaining 5 digits (including 0) can be arranged in the remaining 5 places in \( ^5P_5 \) ways.
\( \therefore \text{Total permutations for six-digit numbers} \)
\( = ^5P_1 \times ^5P_5 \)
\( = 5 \times 5! \)
\( = 5 \times 120 \)
\( = 600 \)
Hence, 600 six-digit numbers can be formed using the given digits.
In simple words: When making numbers, the first spot can't be zero. So, first pick from the non-zero digits. Then, arrange the rest of the digits, including zero, in the remaining spots. Multiply these possibilities together.

Exam Tip: Always remember the special case of '0' when forming numbers. The first digit cannot be zero, which affects the number of choices for the first position.

Question 7. In how many ways can 5 boys and 3 girls be arranged in a row such that all the boys are together?
Answer:
We need to arrange 5 boys and 3 girls in a row so that all the boys remain together.
First, treat the 5 boys as a single block or unit. Let's represent this block as (BBBBB).
Now, we have this one block of boys and the 3 individual girls (G, G, G). So, we are arranging (BBBBB), G, G, G. This gives us a total of 4 units to arrange.
These 4 units can be arranged in \( ^4P_4 \) ways.
\( ^4P_4 = 4! = 24 \)
Next, the 5 boys within their block (BBBBB) can be arranged among themselves in \( ^5P_5 \) ways.
\( ^5P_5 = 5! = 120 \)
To find the total number of arrangements, we multiply the arrangements of the units by the internal arrangements of the boys:
\( \therefore \text{Total permutation} = ^4P_4 \times ^5P_5 \)
\( = 4! \times 5! \)
\( = 24 \times 120 \)
\( = 2880 \)
Hence, 5 boys and 3 girls can be arranged in a row in 2880 ways such that all 5 boys stay together.
In simple words: When things must stay together, group them as one big item. Then, arrange the big item with the others. Finally, arrange the items inside the big item and multiply all the ways together.

Exam Tip: For "things together" problems, consider the group as a single unit first, arrange the units, then arrange items within the group, and multiply the possibilities.

Question 8. There are 7 cages for 7 lions in a zoo. 3 cages out of 7 cages are so small that 3 out of 7 lions cannot fit in it. In how many ways can 7 lions be caged in 7 cages?
Answer:
There are 7 cages in total. 3 cages are small, and 4 cages are big. There are 7 lions in total. 3 lions are big (cannot fit in small cages), and 4 lions are normal-sized (can fit in any cage).
There are 4 big cages and 3 small cages available.

First, we must place the 3 big lions into the 4 big cages, since they cannot fit into the small cages. The number of ways to do this is a permutation of 4 items taken 3 at a time:
Except for 3 small cages, 3 big lions can be caged in the remaining 4 big cages in \( ^4P_3 \) ways.
\( ^4P_3 = 4 \times 3 \times 2 = 24 \)
After placing the 3 big lions in 3 of the big cages, we have \( 4-3=1 \) big cage and 3 small cages remaining. This means we have \( 1+3=4 \) cages left.
We also have \( 7-3=4 \) normal-sized lions remaining.
Now, the remaining 4 lions can be placed in these remaining 4 cages. This can be done in \( ^4P_4 \) ways.
\( ^4P_4 = 4! = 24 \)
\( \therefore \text{Total permutations for caging 7 lions in 7 cages} = ^4P_3 \times ^4P_4 \)
\( = (4 \times 3 \times 2) \times 4! \)
\( = 24 \times 24 \)
\( = 576 \)
Thus, 7 lions can be caged in 7 cages in 576 ways.
In simple words: First, put the large lions into the big cages because they are restricted. Then, put the remaining lions into the remaining cages. Multiply the possibilities from both steps to get the total number of ways.

Exam Tip: In problems with restrictions, always address the restrictions first. Place the items with limitations before placing the unrestricted items.

Question 9. Using all the digits 2, 3, 5, 8, 9, how many numbers greater than 50,000 can be formed?
Answer:
Given digits are 2, 3, 5, 8, 9. We need to form 5-digit numbers using these digits, and the number must be greater than 50,000.
For a 5-digit number to be greater than 50,000, the digit in the first place (ten thousands place) must be 5 or greater. From the given digits, these are 5, 8, 9.
\( \therefore \) Out of the digits 5, 8, and 9, the first digit can be chosen and placed in \( ^3P_1 \) ways.
\( ^3P_1 = 3 \)
After placing one digit in the first position, we have 4 digits remaining. These remaining 4 digits can be arranged in the remaining 4 places in \( ^4P_4 \) ways.
\( ^4P_4 = 4! = 24 \)
\( \therefore \text{Total permutations for numbers greater than } 50,000 = ^3P_1 \times ^4P_4 \)
\( = 3 \times 4! \)
\( = 3 \times 24 \)
\( = 72 \)
Hence, 72 numbers greater than 50,000 can be formed using the given digits.
In simple words: To make a number bigger than 50,000 using these digits, the first digit must be 5, 8, or 9. Choose one for the first spot, then arrange the rest of the digits in the remaining spots. Multiply the options.

Exam Tip: When forming numbers with specific conditions (e.g., greater than a certain value), always consider the restrictions on the leftmost digit first.

Question 10. A person has 5 chocolates of different sizes. These chocolates are to be distributed among 5 children of different ages. If the biggest chocolate is to be given to the youngest child then in how many ways, 5 chocolates can be distributed among 5 children?
Answer:
We have 5 chocolates of different sizes to distribute among 5 children of different ages. A specific condition is given: the biggest chocolate must be given to the youngest child.
This means the distribution of the biggest chocolate to the youngest child is fixed. There is only \( ^1P_1 = 1 \) way for this specific assignment.
After this, there are 4 remaining chocolates and 4 remaining children.
The remaining 4 chocolates can be distributed among the remaining 4 children in \( ^4P_4 \) ways.
\( \therefore \text{Total permutations} = ^1P_1 \times ^4P_4 \)
\( = 1 \times 4! \)
\( = 1 \times 24 \)
\( = 24 \)
Hence, the chocolates can be distributed in 24 different ways under the given condition.
In simple words: Since one chocolate and one child are paired up, we only need to worry about distributing the remaining chocolates among the remaining children. The number of ways to do this is simply the factorial of the remaining items.

Exam Tip: When a specific item is assigned to a specific recipient, that particular arrangement has only one way, effectively reducing the number of items and recipients for further calculation.

Question 11. How many total arrangements can be made using all letters of the following Words?
(1) STATISTICS
(2) BOOKKEEPER
(3) APPEARING
Answer:
(1) For the word STATISTICS:
The word 'STATISTICS' contains 10 letters in total. Let's count the repeated letters:
- 'S' appears 3 times
- 'T' appears 3 times
- 'I' appears 2 times
Here, \( n = 10 \), and the counts of repeated letters are \( p=3, q=3, r=2 \).
The formula for permutations of identical things is \( \frac{n!}{p!q!r!} \).
\( \therefore \text{Total permutation} = \frac{n!}{p!q!r!} \)
\( = \frac{10!}{3!3!2!} \)
\( = \frac{10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{(3 \times 2 \times 1)(3 \times 2 \times 1)(2 \times 1)} \)
\( = \frac{3628800}{6 \times 6 \times 2} \)
\( = \frac{3628800}{72} = 50400 \)

(2) For the word BOOKKEEPER:
The word 'BOOKKEEPER' contains 10 letters in total. Let's count the repeated letters:
- 'B' appears 1 time
- 'O' appears 2 times
- 'K' appears 2 times
- 'E' appears 3 times
- 'P' appears 1 time
- 'R' appears 1 time
Here, \( n = 10 \), and the counts of repeated letters are \( p=3 \) (for E), \( q=2 \) (for O), \( r=2 \) (for K).
\( \therefore \text{Total permutations} = \frac{n!}{p!q!r!} \)
\( = \frac{10!}{3!2!2!} \)
\( = \frac{10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{(3 \times 2 \times 1)(2 \times 1)(2 \times 1)} \)
\( = \frac{3628800}{6 \times 2 \times 2} \)
\( = \frac{3628800}{24} = 151200 \)

(3) For the word APPEARING:
The word 'APPEARING' contains 9 letters in total. Let's count the repeated letters:
- 'A' appears 2 times
- 'P' appears 2 times
- 'E', 'R', 'I', 'N', 'G' each appear 1 time
Here, \( n = 9 \), and the counts of repeated letters are \( p=2 \) (for A), \( q=2 \) (for P).
\( \therefore \text{Total permutations} = \frac{n!}{p!q!} \)
\( = \frac{9!}{2!2!} \)
\( = \frac{9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{(2 \times 1)(2 \times 1)} \)
\( = \frac{362880}{4} = 90720 \)
In simple words: When arranging letters in a word where some letters repeat, you divide the total factorial of all letters by the factorial of each set of repeated letters. This accounts for the identical items.

Exam Tip: Carefully count the total number of letters and the frequency of each repeating letter. A common mistake is miscounting or forgetting a repeated letter.

Question 12. What is the ratio of number of arrangements of all letters of the word ASHOK and GEETA?
Answer:
First, let's find the number of arrangements for the word ASHOK:
The word 'ASHOK' has 5 distinct letters (A, S, H, O, K).
\( \therefore \text{Total permutations of arranging 5 letters of word ASHOK} = ^5P_5 = 5! = 120 \)
Next, let's find the number of arrangements for the word GEETA:
The word 'GEETA' has 5 letters. The letter 'E' is repeated 2 times.
\( \therefore \text{Total permutations of arranging 5 letters of word GEETA, where 'E' is repeated 2 times} = \frac{5!}{2!} = \frac{120}{2} = 60 \)
Now, we calculate the ratio of the number of arrangements of ASHOK to GEETA:
\( \therefore \text{The ratio of number of arrangements of all letters of the word ASHOK and GEETA} = \frac{120}{60} = 2:1 \)
The ratio is 2:1.
In simple words: Find out how many ways you can scramble the letters of ASHOK. Then find out how many ways you can scramble the letters of GEETA, remembering that some letters are the same. Finally, divide the first number by the second.

Exam Tip: When calculating ratios involving permutations, ensure that you correctly calculate the individual permutations, especially if there are repeated letters, before forming the ratio.

Question 13. There are 5 seats in a car including the driver's seat. If 3 out of 10 members in a family know driving then in how many ways, 5 persons out of 10 members can be arranged in the car?
Answer:
There are 5 seats in the car, and one of them is the driver's seat. Out of 10 family members, 3 know how to drive.
First, we need to choose and place a driver. Since only 3 members know driving, the driver's seat can be filled in \( ^3P_1 \) ways.
\( ^3P_1 = 3 \)
After selecting a driver, there are \( 10-1 = 9 \) family members remaining, and \( 5-1 = 4 \) seats remaining (the passenger seats).
The remaining 4 seats need to be filled from the remaining 9 family members. This is a permutation of 9 items taken 4 at a time.
\( ^9P_4 = 9 \times 8 \times 7 \times 6 = 3024 \)
\( \therefore \text{Total permutations for such an arrangement} = ^3P_1 \times ^9P_4 \)
\( = 3 \times (9 \times 8 \times 7 \times 6) \)
\( = 3 \times 3024 \)
\( = 9072 \)
Thus, 5 persons out of 10 members can be arranged in the car in 9072 ways.
In simple words: First, pick someone who can drive for the driver's seat. Then, from the remaining people, pick and arrange them for the remaining empty seats. Multiply these choices together.

Exam Tip: Address special positions (like the driver's seat) and any restrictions (who can drive) first. Then, deal with the remaining people and positions.

Question 14. If the letters of the following words are arranged in dictionary order, what will be the rank of that word?
(1) PINTU
(2) NURI
(3) NIRAL
(4) SUMAN
Answer:
(1) For the word PINTU:
The word PINTU has 5 distinct letters: P, I, N, T, U.
These 5 letters can be arranged in \( ^5P_5 = 5! = 120 \) ways.
To find the rank in dictionary order, we list the letters alphabetically: I, N, P, T, U.
Number of words starting with 'I' (remaining 4 letters arranged in \( ^4P_4 \) ways):
\( ^1P_1 \times ^4P_4 = 1 \times 4! = 24 \)
Number of words starting with 'N' (remaining 4 letters arranged in \( ^4P_4 \) ways):
\( ^1P_1 \times ^4P_4 = 1 \times 4! = 24 \)
Now consider words starting with 'P'. The first letter of PINTU is 'P'. So, we proceed to the second letter.
Words starting with 'PI'. The second letter of PINTU is 'I'.
Now we fix 'P' and 'I' and arrange the remaining 3 letters (N, T, U) alphabetically: N, T, U.
The next letter in PINTU is 'N'. So, we consider words starting with 'PIN'.
Words starting with 'PIN'. The remaining 2 letters (T, U) can be arranged in \( ^2P_2 = 2! = 2 \) ways.
The next letter in PINTU is 'T'. So, we consider words starting with 'PINT'.
Words starting with 'PINT'. The last remaining letter is 'U'. This is PINTU itself.
\( \text{Number of words with P at the first place, I at the second, N at the third, T at the fourth, and U at the fifth place} = ^1P_1 \times ^1P_1 \times ^1P_1 \times ^1P_1 \times ^1P_1 = 1 \), which is the word PINTU itself.
\( \therefore \text{Dictionary order of the word PINTU} = 24 (\text{for I}) + 24 (\text{for N}) + 1 (\text{for PINTU}) = 49 \).

(2) For the word NURI:
The word NURI has 4 distinct letters: N, U, R, I.
These 4 letters can be arranged in \( ^4P_4 = 4! = 24 \) ways.
To find the rank, we list the letters alphabetically: I, N, R, U.
Number of words starting with 'I' (remaining 3 letters arranged in \( ^3P_3 \) ways):
\( ^1P_1 \times ^3P_3 = 1 \times 3! = 6 \)
Now consider words starting with 'N'. The first letter of NURI is 'N'. So, we proceed to the second letter.
Alphabetical order of remaining letters (I, R, U) is I, R, U.
Number of words starting with 'NI' (remaining 2 letters arranged in \( ^2P_2 \) ways):
\( ^1P_1 \times ^1P_1 \times ^2P_2 = 1 \times 1 \times 2! = 2 \)
Number of words starting with 'NR' (remaining 2 letters arranged in \( ^2P_2 \) ways):
\( ^1P_1 \times ^1P_1 \times ^2P_2 = 1 \times 1 \times 2! = 2 \)
Now consider words starting with 'NU'. The second letter of NURI is 'U'.
Alphabetical order of remaining letters (I, R) is I, R.
Number of words starting with 'NUI' (remaining 1 letter arranged in \( ^1P_1 \) ways):
\( ^1P_1 \times ^1P_1 \times ^1P_1 \times ^1P_1 = 1 \)
Now consider words starting with 'NUR'. The third letter of NURI is 'R'.
The last remaining letter is 'I'. This forms NURI itself.
\( \text{Number of words with N at the first place, U at the second place, R at the third place and I at the fourth place} = 1 \), which is the word NURI itself.
\( \therefore \text{Dictionary order of the word NURI} = 6 (\text{for I}) + 2 (\text{for NI}) + 2 (\text{for NR}) + 1 (\text{for NUI}) + 1 (\text{for NURI}) = 12 \).

(3) For the word NIRAL:
The word NIRAL has 5 distinct letters: N, I, R, A, L.
These 5 letters can be arranged in \( ^5P_5 = 5! = 120 \) ways.
To find the rank, we list the letters alphabetically: A, I, L, N, R.
Number of words starting with 'A' (remaining 4 letters arranged in \( ^4P_4 \) ways):
\( ^1P_1 \times ^4P_4 = 1 \times 4! = 24 \)
Number of words starting with 'I' (remaining 4 letters arranged in \( ^4P_4 \) ways):
\( ^1P_1 \times ^4P_4 = 1 \times 4! = 24 \)
Number of words starting with 'L' (remaining 4 letters arranged in \( ^4P_4 \) ways):
\( ^1P_1 \times ^4P_4 = 1 \times 4! = 24 \)
Now consider words starting with 'N'. The first letter of NIRAL is 'N'. So, we proceed to the second letter.
Alphabetical order of remaining letters (A, I, L, R) is A, I, L, R.
Number of words starting with 'NA' (remaining 3 letters arranged in \( ^3P_3 \) ways):
\( ^1P_1 \times ^1P_1 \times ^3P_3 = 1 \times 1 \times 3! = 6 \)
Number of words starting with 'NI'. The second letter of NIRAL is 'I'. So, we proceed to the third letter.
Alphabetical order of remaining letters (A, L, R) is A, L, R.
Number of words starting with 'NIA' (remaining 2 letters arranged in \( ^2P_2 \) ways):
\( ^1P_1 \times ^1P_1 \times ^1P_1 \times ^2P_2 = 1 \times 1 \times 1 \times 2! = 2 \)
Number of words starting with 'NIL' (remaining 2 letters arranged in \( ^2P_2 \) ways):
\( ^1P_1 \times ^1P_1 \times ^1P_1 \times ^2P_2 = 1 \times 1 \times 1 \times 2! = 2 \)
Now consider words starting with 'NIR'. The third letter of NIRAL is 'R'. So, we proceed to the fourth letter.
Alphabetical order of remaining letters (A, L) is A, L.
The next letter in NIRAL is 'A'. So, we consider words starting with 'NIRA'.
Words starting with 'NIRA'. The last remaining letter is 'L'. This forms NIRAL itself.
\( \text{Number of words with N at the first place, I at the second place, R at the third place, A at the fourth place and L at the fifth place} = 1 \), which is the word NIRAL.
\( \therefore \text{Dictionary order of the word NIRAL} = 24+24+24+6+2+2+1 = 83 \).

(4) For the word SUMAN:
The word SUMAN has 5 distinct letters: S, U, M, A, N.
These 5 letters can be arranged in \( ^5P_5 = 5! = 120 \) ways.
To find the rank, we list the letters alphabetically: A, M, N, S, U.
Number of words starting with 'A' (remaining 4 letters arranged in \( ^4P_4 \) ways):
\( ^1P_1 \times ^4P_4 = 1 \times 4! = 24 \)
Number of words starting with 'M' (remaining 4 letters arranged in \( ^4P_4 \) ways):
\( ^1P_1 \times ^4P_4 = 1 \times 4! = 24 \)
Number of words starting with 'N' (remaining 4 letters arranged in \( ^4P_4 \) ways):
\( ^1P_1 \times ^4P_4 = 1 \times 4! = 24 \)
Now consider words starting with 'S'. The first letter of SUMAN is 'S'. So, we proceed to the second letter.
Alphabetical order of remaining letters (A, M, N, U) is A, M, N, U.
Number of words starting with 'SA' (remaining 3 letters arranged in \( ^3P_3 \) ways):
\( ^1P_1 \times ^1P_1 \times ^3P_3 = 1 \times 1 \times 3! = 6 \)
Number of words starting with 'SM' (remaining 3 letters arranged in \( ^3P_3 \) ways):
\( ^1P_1 \times ^1P_1 \times ^3P_3 = 1 \times 1 \times 3! = 6 \)
Number of words starting with 'SN' (remaining 3 letters arranged in \( ^3P_3 \) ways):
\( ^1P_1 \times ^1P_1 \times ^3P_3 = 1 \times 1 \times 3! = 6 \)
Now consider words starting with 'SU'. The second letter of SUMAN is 'U'. So, we proceed to the third letter.
Alphabetical order of remaining letters (A, M, N) is A, M, N.
The next letter in SUMAN is 'M'. So, we consider words starting with 'SUM'.
Words starting with 'SUM'. The remaining 2 letters (A, N) can be arranged in \( ^2P_2 = 2! = 2 \) ways.
The next letter in SUMAN is 'A'. So, we consider words starting with 'SUMA'.
Words starting with 'SUMA'. The last remaining letter is 'N'. This forms SUMAN itself.
\( \text{Number of words with S at the first place, U at the second place, M at the third place, A at the fourth place and N at the fifth place} = 1 \), which is the word SUMAN.
\( \therefore \text{Dictionary order of the word SUMAN} = 24+24+24+6+6+6+1 = 93 \).
In simple words: To find a word's dictionary rank, count all words that come before it alphabetically by fixing the letters one by one from left to right. Sum up the counts of all possibilities for each position.

Exam Tip: Systematically list the alphabetical order of letters. For each position, count all permutations of letters that come before the target word's letter. This method ensures accuracy in calculating dictionary rank.

Question 15. How many arrangements of the letters of the word SHLOKA can be made such that all vowels are together?
Answer:
The word 'SHLOKA' has 6 letters. The vowels in the word are O and A. The consonants are S, H, L, K.
We need to arrange the letters so that all vowels are together. To do this, we treat the group of vowels (OA) as a single unit.
Now, we have 5 units to arrange: (OA), S, H, L, K.
These 5 units can be arranged in \( ^5P_5 \) ways.
\( ^5P_5 = 5! = 120 \)
Within the vowel unit (OA), the two vowels can be arranged among themselves in \( ^2P_2 \) ways.
\( ^2P_2 = 2! = 2 \)
To find the total number of arrangements, we multiply the arrangements of the units by the internal arrangements of the vowels:
\( \therefore \text{Total permutations} = ^5P_5 \times ^2P_2 \)
\( = 5! \times 2! \)
\( = 120 \times 2 \)
\( = 240 \)
Thus, 240 arrangements of the letters of SHLOKA can be made such that all vowels are together.
In simple words: Group the vowels into one block. Then, arrange this block with the other letters. Finally, arrange the vowels inside their block. Multiply these results to get the total number of arrangements.

Exam Tip: When elements must stay together, treat them as a single entity for the initial arrangement. Then, consider the internal arrangements of that entity and multiply the results.

Question 16. 7 speakers A, B, C, D, E, F and G are invited to deliver a speech in a program. Speakers have to deliver speech one after the other. In how many ways, speeches of 7 speakers can be arranged if the speaker B has to deliver his speech immediately after the speaker A?
Answer:
There are 7 speakers: A, B, C, D, E, F, G. The condition is that speaker B must deliver his speech immediately after speaker A.
This means we treat 'AB' as a single block or unit. So, the order 'AB' is fixed.
Now, we need to arrange this 'AB' block along with the remaining 5 speakers (C, D, E, F, G). This gives us a total of 6 units to arrange: (AB), C, D, E, F, G.
The number of ways to arrange these 6 distinct units is \( ^6P_6 \).
\( \therefore \text{Total permutations} \)
\( = ^6P_6 \)
\( = 6! \)
\( = 6 \times 5 \times 4 \times 3 \times 2 \times 1 \)
\( = 720 \)
Thus, the speeches can be arranged in 720 ways under the given condition.
In simple words: Since B must follow A right away, treat 'AB' like one person. Then, arrange this 'AB' person with the other 5 speakers. This is like arranging 6 people, which is 6 factorial.

Exam Tip: If two specific items must always occur in a fixed order, treat them as a single combined unit. This reduces the total number of items to be arranged by one.