CBSE Class 11 Mathematics Permutations And Combinations Worksheet Set 05

CBSE Class 11 Mathematics Worksheet - Permutations and Combinations (4). Students can download these worksheets and practice them. This will help them to get better marks in examinations. Also refer to other worksheets for the same chapter and other subjects too. Use them for better understanding of the subjects.

Question. Find the number of ways in which six ‘+’ and four ‘-’ signs can be arranged in a line such that no 2 signs ‘-’ occur together?
Answer: We have:
Number of '+' signs \( = 6 \)
Number of '-' signs \( = 4 \)

First, we arrange the 6 '+' signs in a row. Since they are identical, they can be arranged in only 1 way:
\( \frac{6!}{6!} = 1 \text{ way} \)

Arranging the 6 '+' signs creates 7 available gaps (indicated by underscores) where the '-' signs can be placed so that no two '-' signs are adjacent:
\( \text{_ + _ + _ + _ + _ + _ + _} \)

There are 7 places available for the 4 identical '-' signs, which can be selected and arranged in:
\( {}^7C_4 = \frac{7!}{4!3!} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35 \text{ ways} \)

\( \therefore \text{Required number of ways} = 35 \times 1 = 35 \text{ ans.} \)
In simple words: Arrange the '+' signs first to create gaps, then place the '-' signs into these gaps so no two of them are next to each other.

Exam Tip: Remember that since all '+' and '-' signs are identical, we use combinations \( \left({}^7C_4\right) \) rather than permutations to place them in the gaps.

Question. Find the number of ways in which we can choose committee from four men & six women so that the committee includes at least two men and exactly twice as many women as men?
Answer: Let \( M \) be the number of men and \( W \) be the number of women selected. Since the committee must have exactly twice as many women as men, we have \( W = 2M \).
Given there are 4 men and 6 women available, and the committee must include at least 2 men, we consider the following cases:

- Case 1: 2 Men and 4 Women (\( M = 2, W = 4 \))
Selecting 2 men out of 4 and 4 women out of 6 can be done in:
\( {}^4C_2 \times {}^6C_4 = 6 \times 15 = 90 \text{ ways} \)

- Case 2: 3 Men and 6 Women (\( M = 3, W = 6 \))
Selecting 3 men out of 4 and 6 women out of 6 can be done in:
\( {}^4C_3 \times {}^6C_6 = 4 \times 1 = 4 \text{ ways} \)

- Case 3: 4 Men and 8 Women (\( M = 4, W = 8 \))
This is not possible because only 6 women are available in total.

\( \therefore \text{Total required number of ways} = 90 + 4 = 94 \text{ ans.} \)
In simple words: Find the different ways to choose 2 men and 4 women, and 3 men and 6 women, then add those possibilities together.

Exam Tip: Be careful to check the upper limit of available people; since there are only 6 women, we cannot have a case with 4 men and 8 women.

Question. Ten different letters of alphabet are given words with five letters are formed from these given letters. Then find the number of words which have at least one letter repeated?
Answer: We use the principle of subtraction to find the required count:

1. The total number of 5-letter words that can be formed from 10 letters with repetition allowed is:
\( 10 \times 10 \times 10 \times 10 \times 10 = 10^5 = 100,000 \)

2. The number of 5-letter words with no repeating letters is:
\( 10 \times 9 \times 8 \times 7 \times 6 = 30,240 \)

3. Therefore, the number of words with at least one letter repeated is:
\( \text{Required words} = \text{Total words} - \text{Words with no repetition} \)
\( = 10^5 - 30,240 \)
\( = 100,000 - 30,240 = 69,760 \text{ ans.} \)
In simple words: To find the words with at least one repeated letter, subtract the number of words with all distinct letters from the total possible words.

Exam Tip: The "total minus none" method is almost always the fastest way to solve "at least one" probability or permutation questions.

Question. There are 10 lamps in a hall. Each one of them can be switched on independently. Find the number of ways in which the hall can be illuminated?
Answer: Each of the 10 lamps has 2 possible states: it can be either switched ON or switched OFF.

The total number of ways to set the states of all 10 lamps is:
\( 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 = 2^{10} = 1024 \text{ ways} \)

Out of these 1024 combinations, there is exactly 1 way in which all 10 lamps are switched OFF simultaneously (in which case the hall is not illuminated).

\( \therefore \text{Required number of ways to illuminate the hall} = 2^{10} - 1 \)
\( = 1024 - 1 = 1023 \text{ ans.} \)
In simple words: Each lamp has 2 options (on or off). We calculate the total combinations for 10 lamps and subtract the single case where all of them are off.

Exam Tip: Always read the question carefully; "illuminated" means at least one lamp must be on, which is why we must subtract 1 from \( 2^{10} \).

Question. A five digit number divisible by 3 is to be formed using the numbers 0,1,2,3,4,5 without repetitions. Find total no. of ways this can be done?
Answer: A number is divisible by 3 if the sum of its digits is divisible by 3.
The sum of all 6 given digits \( \{0, 1, 2, 3, 4, 5\} \) is \( 0 + 1 + 2 + 3 + 4 + 5 = 15 \) (which is divisible by 3).
To form a 5-digit number, we must select 5 digits. The sum of the chosen 5 digits will be divisible by 3 if we exclude a digit that is a multiple of 3 (either 0 or 3):

- Case 1: Excluding 3 (using digits 0, 1, 2, 4, 5):
The ten-thousands place cannot be 0, so it can be filled in 4 ways. The remaining four positions can be filled in \( 4 \), \( 3 \), \( 2 \), and \( 1 \) ways respectively.
Number of ways \( = 4 \times 4 \times 3 \times 2 \times 1 = 96 \)

- Case 2: Excluding 0 (using digits 1, 2, 3, 4, 5):
Since there is no 0, the 5 positions can be filled in \( 5! \) ways.
Number of ways \( = 5 \times 4 \times 3 \times 2 \times 1 = 120 \)

\( \therefore \text{Total required number of ways} = 96 + 120 = 216 \text{ ans.} \)
In simple words: A number is divisible by 3 if its digits add up to a multiple of 3. Find the sets of 5 digits that satisfy this, and calculate the arrangements for each set.

Exam Tip: Do not forget that when 0 is included in the chosen digits, it cannot be placed in the first position of a 5-digit number.

Question. The number 5-digit telephone number having at least one of their digit is repeated?
Answer: We use the subtraction method:

1. The total number of 5-digit telephone numbers that can be formed using digits \( 0-9 \) (where the first digit can be 0) is:
\( 10^5 = 100,000 \)

2. The number of 5-digit numbers with no repeating digits is:
\( 10 \times 9 \times 8 \times 7 \times 6 = 30,240 \)

3. Therefore, the number of 5-digit telephone numbers with at least one digit repeated is:
\( \text{Required numbers} = 100,000 - 30,240 = 69,760 \text{ ans.} \)
In simple words: Subtract the number of 5-digit numbers with no repeating digits from the total possible 5-digit numbers to find those with at least one repetition.

Exam Tip: Unless specified otherwise, assume telephone numbers can start with 0, meaning there are 10 choices for the first digit.

Question. In a football championship, 153 matches were played. Every 2 teams played one match with each other. The number of teams participating in the championship?
Answer: Let \( n \) be the number of teams participating in the championship.
Since every match is played between 2 teams, the total number of matches played is given by \( {}^nC_2 \):
\( {}^nC_2 = 153 \)
\( \Rightarrow \frac{n(n-1)}{2} = 153 \)
\( \Rightarrow n(n-1) = 306 \)
\( \Rightarrow n^2 - n - 306 = 0 \)
\( \Rightarrow (n - 18)(n + 17) = 0 \)
\( \Rightarrow n = 18 \text{ or } n = -17 \)

Since the number of teams must be a positive integer, we reject \( n = -17 \).
\( \therefore \) The number of participating teams is \( 18 \) ans.
In simple words: Each match is played between 2 teams. We solve the equation \( {}^nC_2 = 153 \) to find the total number of teams, \( n \).

Exam Tip: When solving quadratic equations in permutation problems, discard negative roots since the number of teams must be a positive integer.

Question. A lady gives a dinner party for 6 guests. Find the number of ways in which they may be selected from among 10 friends if 2 of the friend will not attend the party together?
Answer: We use the formula:
\( \text{Required ways} = \text{Total ways to select guests} - \text{Ways where the 2 friends attend together} \)

1. Total ways to select 6 guests from 10 friends is:
\( {}^{10}C_6 = {}^{10}C_4 = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = 210 \)

2. If the 2 particular friends attend together, we must select the remaining 4 guests from the other 8 friends:
\( {}^2C_2 \times {}^8C_4 = 1 \times 70 = 70 \)

3. Therefore, the number of ways in which the 2 friends do not attend together is:
\( \text{Required ways} = 210 - 70 = 140 \text{ ans.} \)
In simple words: Calculate the total ways to choose 6 guests from 10, then subtract the ways where the 2 particular friends are selected together.

Exam Tip: Grouping elements that cannot be together into a "together" case and subtracting it from the total is a highly reliable strategy.

Question. We wish to select 6 persons from 8, but if the person A is chosen then B must be chosen. In how many ways can the selection be made?
Answer: We divide the selection process into two mutually exclusive cases:

- Case 1: When person A is chosen:
Since A is chosen, B must also be chosen. Thus, 2 persons (A and B) are already selected. We now select the remaining 4 persons from the remaining 6 persons:
\( \text{Ways} = {}^6C_4 = {}^6C_2 = \frac{6 \times 5}{2 \times 1} = 15 \)

- Case 2: When person A is not chosen:
Since A is not selected, we can choose any 6 persons from the remaining 7 persons (which may or may not include B):
\( \text{Ways} = {}^7C_6 = 7 \)

\( \therefore \text{Total required number of ways} = 15 + 7 = 22 \text{ ans.} \)
In simple words: Solve by splitting into two cases: one where person A is chosen (which forces B to be chosen too), and one where A is not chosen.

Exam Tip: Break conditional selection problems into disjoint cases (A is included vs. A is excluded) to make the calculations straightforward and error-free.

Question. Find the maximum number of points of intersection of 8 straight lines in a plane?
Answer: Two straight lines intersect at exactly 1 point. Therefore, the maximum number of points of intersection is equal to the number of ways to choose 2 lines from the given 8 lines:
\( \text{Maximum points} = {}^8C_2 = \frac{8 \times 7}{2 \times 1} = 28 \text{ ans.} \)
In simple words: Since any 2 lines intersect at exactly 1 point, the maximum number of intersection points is the number of ways to choose 2 lines from 8.

Exam Tip: Maximum intersection occurs when no two lines are parallel and no three lines intersect at the same point.

Question. In how many ways can the letters of the word PERMUTATIONS be arranged, if there are always 4 letters between P and S?
Answer: The word "PERMUTATIONS" has 12 letters, where 'T' repeats twice.

- First, we find the number of ways to place P and S such that there are exactly 4 letters between them. The possible positional pairs are:
\( (1, 6), (2, 7), (3, 8), (4, 9), (5, 10), (6, 11), (7, 12) \) (7 positions).
Since P and S can interchange places, there are \( 7 \times 2 = 14 \) ways to arrange P and S.

- Next, we arrange the remaining 10 letters (with 'T' repeating twice) in the remaining 10 places:
\( \text{Arrangements} = \frac{10!}{2!} = 1,814,400 \text{ ways} \)

\( \therefore \text{Total required number of ways} = 14 \times 1,814,400 = 25,401,600 \text{ ans.} \br /> In simple words: Place P and S in positions with exactly 4 gaps between them, and then arrange the remaining 10 letters in the other positions.

Exam Tip: Don't forget to divide by \( 2! \) for the repeating 'T's in the word "PERMUTATIONS" to get the correct final count.