OP Malhotra Class 11 Maths Solutions Chapter 12 Permutations and Combinations Exercise 12 (D)

Question 1. Find the value of:
(i) \( ^5C_2 \)
(ii) \( ^{10}C_4 \)
(iii) \( ^{50}C_{47} \)
Answer:
(i) To find the value of \( ^5C_2 \):
\( ^5C_2 = \frac{5!}{ (5-2)!2! } = \frac{5!}{3!2!} \)
\( = \frac{5 \times 4 \times 3!}{3! \times 2 \times 1} \)
\( = \frac{5 \times 4}{2} \)
\( = 10 \)

(ii) To find the value of \( ^{10}C_4 \):
\( ^{10}C_4 = \frac{10!}{ (10-4)!4! } = \frac{10!}{6!4!} \)
\( = \frac{10 \times 9 \times 8 \times 7 \times 6!}{6! \times 4 \times 3 \times 2 \times 1} \)
\( = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} \)
\( = 10 \times 3 \times 7 \)
\( = 210 \)

(iii) To find the value of \( ^{50}C_{47} \):
\( ^{50}C_{47} = \frac{50!}{ (50-47)!47! } = \frac{50!}{3!47!} \)
\( = \frac{50 \times 49 \times 48 \times 47!}{3 \times 2 \times 1 \times 47!} \)
\( = \frac{50 \times 49 \times 48}{3 \times 2 \times 1} \)
\( = 50 \times 49 \times 8 \)
\( = 19600 \)
In simple words: Combinations tell us how many ways we can pick items from a group without caring about their order. We use the formula \( ^nC_r = \frac{n!}{r!(n-r)!} \) to calculate these values.

🎯 Exam Tip: Remember that \( ^nC_r = ^nC_{n-r} \). This property can simplify calculations, for example, \( ^{50}C_{47} \) is the same as \( ^{50}C_{3} \), which might be easier to compute. Always try to simplify before multiplying large numbers.

Question 2. Evaluate :
(i) C(15, 14)
(ii) C(8, 5)
(iii) \( ^{11}C_2 \)
Answer:
(i) To evaluate \( C(15, 14) \) which is \( ^{15}C_{14} \):
\( ^{15}C_{14} = ^{15}C_{(15-14)} \)
\( = ^{15}C_1 \)
\( = \frac{15!}{1!(15-1)!} \)
\( = \frac{15!}{1!14!} \)
\( = \frac{15 \times 14!}{1 \times 14!} \)
\( = 15 \)

(ii) To evaluate \( C(8, 5) \) which is \( ^8C_5 \):
\( ^8C_5 = \frac{8!}{5!(8-5)!} \)
\( = \frac{8!}{5!3!} \)
\( = \frac{8 \times 7 \times 6 \times 5!}{5! \times 3 \times 2 \times 1} \)
\( = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} \)
\( = 8 \times 7 \)
\( = 56 \)

(iii) To evaluate \( ^{11}C_2 \):
\( ^{11}C_2 = \frac{11!}{2!(11-2)!} \)
\( = \frac{11!}{2!9!} \)
\( = \frac{11 \times 10 \times 9!}{2 \times 1 \times 9!} \)
\( = \frac{11 \times 10}{2} \)
\( = 11 \times 5 \)
\( = 55 \)
In simple words: We are calculating the number of ways to choose a small group from a larger set. For example, if you have 8 different items, there are 56 ways to pick any 5 of them.

🎯 Exam Tip: Use the property \( ^nC_r = ^nC_{n-r} \) especially when \( r \) is a large number close to \( n \), as it often simplifies calculations significantly (e.g., \( ^{15}C_{14} \) becomes \( ^{15}C_1 \)).

Question 3. Evaluate:
(i) C(19, 17) + C(19, 18)
(ii) C(31, 26) – C(30, 26)
Answer:
(i) To evaluate \( C(19, 17) + C(19, 18) \):
This can be written as \( ^{19}C_{17} + ^{19}C_{18} \).
Using the identity \( ^nC_r + ^nC_{r-1} = ^{n+1}C_r \), we can write:
\( ^{19}C_{17} + ^{19}C_{18} = ^{19}C_{17} + ^{19}C_{19-1} \). This identity applies if the second term is \( ^{19}C_{17+1} \) or \( ^{19}C_{18} \) where \( r=18 \).
The formula is \( ^nC_r + ^nC_{r+1} = ^{n+1}C_{r+1} \). Here \( n=19 \), \( r=17 \). So \( ^{19}C_{17} + ^{19}C_{18} = ^{19+1}C_{18} = ^{20}C_{18} \).
Now, calculate \( ^{20}C_{18} \):
\( ^{20}C_{18} = ^{20}C_{(20-18)} = ^{20}C_2 \)
\( = \frac{20!}{2!(20-2)!} = \frac{20!}{2!18!} \)
\( = \frac{20 \times 19 \times 18!}{2 \times 1 \times 18!} \)
\( = \frac{20 \times 19}{2} \)
\( = 10 \times 19 \)
\( = 190 \)
Alternatively, by direct calculation as in the source:
\( ^{19}C_{17} = \frac{19 \times 18}{2 \times 1} = 19 \times 9 = 171 \)
\( ^{19}C_{18} = ^{19}C_1 = 19 \)
\( 171 + 19 = 190 \)

(ii) To evaluate \( C(31, 26) – C(30, 26) \):
This is \( ^{31}C_{26} – ^{30}C_{26} \).
We know the identity \( ^nC_r = ^{n-1}C_r + ^{n-1}C_{r-1} \).
This can be rearranged as \( ^nC_r - ^{n-1}C_r = ^{n-1}C_{r-1} \).
Here, \( n = 31 \) and \( r = 26 \). So, \( ^{31}C_{26} – ^{30}C_{26} = ^{30}C_{25} \).
Let's calculate \( ^{30}C_{25} \):
\( ^{30}C_{25} = ^{30}C_{(30-25)} = ^{30}C_5 \)
\( = \frac{30!}{5!(30-5)!} = \frac{30!}{5!25!} \)
\( = \frac{30 \times 29 \times 28 \times 27 \times 26 \times 25!}{5 \times 4 \times 3 \times 2 \times 1 \times 25!} \)
\( = \frac{30 \times 29 \times 28 \times 27 \times 26}{5 \times 4 \times 3 \times 2 \times 1} \)
\( = 6 \times 29 \times 7 \times 9 \times 13 \)
\( = 142506 \)
In simple words: We used special rules for combinations to solve these problems quickly. The first problem used a rule that lets you add two combinations, and the second problem used a rule that lets you subtract them.

🎯 Exam Tip: Familiarize yourself with Pascal's identity: \( ^nC_r + ^nC_{r+1} = ^{n+1}C_{r+1} \) and its variations. These identities are very helpful for simplifying calculations and often save a lot of time in exams.

Question 4. If \( ^4C_2 = n \cdot ^4C_2 \), find n.
Answer:
Given the equation: \( ^4C_2 = n \cdot ^4C_2 \).
First, calculate \( ^4C_2 \):
\( ^4C_2 = \frac{4!}{2!(4-2)!} = \frac{4!}{2!2!} \)
\( = \frac{4 \times 3 \times 2 \times 1}{(2 \times 1)(2 \times 1)} = \frac{24}{4} = 6 \)
So, the equation becomes \( 6 = n \times 6 \).
To find \( n \), divide both sides by 6:
\( n = \frac{6}{6} \)
\( n = 1 \)
In simple words: The problem states that a number is equal to itself multiplied by 'n'. This can only be true if 'n' is 1, otherwise the numbers would not be equal.

🎯 Exam Tip: When you see the same term on both sides of an equation, you can often cancel it out, but be careful if the term could be zero. In this case, \( ^4C_2 \) is 6, which is not zero, so cancellation is safe.

Question 5. If \( ^nC_4 = ^nC_6 \), find n.
Answer:
Given the equation: \( ^nC_4 = ^nC_6 \).
We use the property of combinations that states if \( ^nC_r = ^nC_s \), then either \( r = s \) or \( r + s = n \).
In this case, \( r = 4 \) and \( s = 6 \).
Since \( r \neq s \) (4 is not equal to 6), the second condition must be true.
Therefore, \( r + s = n \).
Substitute the values of \( r \) and \( s \):
\( 4 + 6 = n \)
\( n = 10 \)
In simple words: If you can choose 4 items from a group in the same number of ways as choosing 6 items from the same group, it means the total number of items must be 4 plus 6, which is 10.

🎯 Exam Tip: Always remember the property \( ^nC_r = ^nC_s \implies r=s \text{ or } r+s=n \). This property is fundamental for solving many combination-based equations efficiently without expanding factorials.

Question 6. If \( C(2n, 3) : C(n, 2) = 12 : 1 \), find n.
Answer:
Given the ratio: \( C(2n, 3) : C(n, 2) = 12 : 1 \).
This can be written as \( \frac{^{2n}C_3}{^nC_2} = \frac{12}{1} \).
First, expand the combination terms:
\( ^{2n}C_3 = \frac{(2n)!}{3!(2n-3)!} = \frac{2n(2n-1)(2n-2)(2n-3)!}{3 \times 2 \times 1 \times (2n-3)!} = \frac{2n(2n-1)(2n-2)}{6} \)
\( ^nC_2 = \frac{n!}{2!(n-2)!} = \frac{n(n-1)(n-2)!}{2 \times 1 \times (n-2)!} = \frac{n(n-1)}{2} \)
Now substitute these into the ratio:
\( \frac{\frac{2n(2n-1)(2n-2)}{6}}{\frac{n(n-1)}{2}} = 12 \)
\( \implies \frac{2n(2n-1)(2n-2)}{6} \times \frac{2}{n(n-1)} = 12 \)
\( \implies \frac{2n(2n-1)2(n-1)}{6} \times \frac{2}{n(n-1)} = 12 \)
Cancel \( n \) and \( (n-1) \) (assuming \( n \neq 0 \) and \( n \neq 1 \)):
\( \implies \frac{4(2n-1)}{3} = 12 \)
\( \implies 4(2n-1) = 12 \times 3 \)
\( \implies 4(2n-1) = 36 \)
\( \implies 2n-1 = \frac{36}{4} \)
\( \implies 2n-1 = 9 \)
\( \implies 2n = 10 \)
\( \implies n = 5 \)
We must check if \( n=0 \) or \( n=1 \) are valid. For combinations \( ^nC_r \), we need \( n \ge r \).
For \( ^{2n}C_3 \), if \( n=0 \), \( ^0C_3 \) is undefined. If \( n=1 \), \( ^2C_3 \) is undefined. So \( n=0 \) and \( n=1 \) do not satisfy the original equation.
Thus, the only valid solution is \( n = 5 \).
In simple words: We used the combination formula and then simplified the ratio to find the value of 'n'. We also made sure that the value of 'n' worked for the combination rule (n must be greater than or equal to r).

🎯 Exam Tip: When solving combination equations, always expand the factorial terms carefully and simplify before multiplying. Also, remember to check any potential values of 'n' (like 0 or 1 in this case) against the conditions for combinations to be defined (\( n \ge r \)).

Question 7. If \( ^nC_r : ^nC_{r+1} = 1 : 2 \) and \( ^nC_{r+1} : ^nC_{r+2} = 2 : 3 \), determine the values of n and r.
Answer:
Given the first ratio: \( \frac{^nC_r}{^nC_{r+1}} = \frac{1}{2} \).
We know that \( \frac{^nC_r}{^nC_{r+1}} = \frac{r+1}{n-r} \).
So, \( \frac{r+1}{n-r} = \frac{1}{2} \)
Cross-multiply:
\( 2(r+1) = n-r \)
\( 2r + 2 = n - r \)
\( n - 3r = 2 \) ...(1)

Given the second ratio: \( \frac{^nC_{r+1}}{^nC_{r+2}} = \frac{2}{3} \).
Using the same identity, \( \frac{^nC_{r+1}}{^nC_{r+2}} = \frac{(r+2)+1}{n-(r+1)} = \frac{r+2}{n-r-1} \).
So, \( \frac{r+2}{n-(r+1)} = \frac{2}{3} \)
Cross-multiply:
\( 3(r+2) = 2(n-r-1) \)
\( 3r + 6 = 2n - 2r - 2 \)
\( 2n - 5r = 8 \) ...(2)

Now we have a system of two linear equations:
1) \( n - 3r = 2 \)
2) \( 2n - 5r = 8 \)
From equation (1), express \( n \) in terms of \( r \):
\( n = 3r + 2 \)
Substitute this expression for \( n \) into equation (2):
\( 2(3r+2) - 5r = 8 \)
\( 6r + 4 - 5r = 8 \)
\( r + 4 = 8 \)
\( r = 4 \)
Now substitute the value of \( r=4 \) back into the expression for \( n \):
\( n = 3(4) + 2 \)
\( n = 12 + 2 \)
\( n = 14 \)
So, the values are \( n=14 \) and \( r=4 \).
In simple words: We used a quick math rule that helps simplify ratios of combinations into simpler equations. By solving these two simple equations together, we found the values for 'n' and 'r' that satisfy both conditions.

🎯 Exam Tip: Remember the useful identity for ratios of consecutive combinations: \( \frac{^nC_r}{^nC_{r+1}} = \frac{r+1}{n-r} \). This identity is crucial for solving problems involving ratios of combinations quickly and efficiently by converting them into linear equations.

Question 8. If \( C(n, 10) = C(n, 12) \), determine \( C(n, 5) \).
Answer:
Given that \( C(n, 10) = C(n, 12) \). This can be written as \( ^nC_{10} = ^nC_{12} \).
Using the property of combinations that states if \( ^nC_r = ^nC_s \), then either \( r = s \) or \( r + s = n \).
Here, \( r = 10 \) and \( s = 12 \).
Since \( r \neq s \) (10 is not equal to 12), the second condition must be true:
\( r + s = n \)
Substitute the values of \( r \) and \( s \):
\( 10 + 12 = n \)
\( n = 22 \)
Now we need to determine \( C(n, 5) \), which is \( ^nC_5 \). Since \( n=22 \), we need to calculate \( ^{22}C_5 \).
\( ^{22}C_5 = \frac{22!}{5!(22-5)!} = \frac{22!}{5!17!} \)
\( = \frac{22 \times 21 \times 20 \times 19 \times 18 \times 17!}{5 \times 4 \times 3 \times 2 \times 1 \times 17!} \)
\( = \frac{22 \times 21 \times 20 \times 19 \times 18}{5 \times 4 \times 3 \times 2 \times 1} \)
\( = 22 \times (21/3) \times (20/(5 \times 4)) \times 19 \times (18/2) \)
\( = 22 \times 7 \times 1 \times 19 \times 9 \)
\( = 26334 \)
In simple words: First, we found the total number of items 'n' using a special rule for combinations where two different choices give the same result. Once we found 'n', we then calculated how many ways we could choose 5 items from that total.

🎯 Exam Tip: This question combines two important concepts: using the property \( ^nC_r = ^nC_s \implies r+s=n \) to find \( n \), and then using the standard combination formula to calculate a specific combination. Make sure you are proficient in both.

Question 9. If \( C(2n, r) = C(2n, r + 2) \), find r in terms of n.
Answer:
Given the equation: \( C(2n, r) = C(2n, r + 2) \). This can be written as \( ^{2n}C_r = ^{2n}C_{r+2} \).
Using the property of combinations that states if \( ^N C_R = ^N C_S \), then either \( R = S \) or \( R + S = N \).
In this case, \( N = 2n \), \( R = r \), and \( S = r+2 \).
Since \( R \neq S \) (as \( r \neq r+2 \)), the second condition must be true:
\( R + S = N \)
Substitute the values:
\( r + (r+2) = 2n \)
\( 2r + 2 = 2n \)
Divide the entire equation by 2:
\( r + 1 = n \)
So, \( r = n - 1 \).
In simple words: This problem uses a simple rule of combinations. If choosing 'r' items from a group gives the same number of ways as choosing 'r+2' items from the same group, then the total number of items must be equal to 'r' plus 'r+2'. We used this to find 'r' in terms of 'n'.

🎯 Exam Tip: When applying the \( ^nC_r = ^nC_s \) property, make sure to correctly identify the "total items" (N) and the "chosen items" (R and S). In this case, \( 2n \) is the total number of items, not just \( n \).

Question 10. The value of \( ^{50}C_4 + \sum_{r=1}^{6} {^{56-r}C_3} \) is
(a) \( ^{55}C_4 \)
(b) \( ^{55}C_3 \)
(c) \( ^{56}C_3 \)
(d) \( ^{56}C_4 \)
Answer: (d) \( ^{56}C_4 \)
We need to evaluate \( ^{50}C_4 + \sum_{r=1}^{6} {^{56-r}C_3} \).
Let's expand the summation: \[ \sum_{r=1}^{6} {^{56-r}C_3} = ^{56-1}C_3 + ^{56-2}C_3 + ^{56-3}C_3 + ^{56-4}C_3 + ^{56-5}C_3 + ^{56-6}C_3 \] \[ = ^{55}C_3 + ^{54}C_3 + ^{53}C_3 + ^{52}C_3 + ^{51}C_3 + ^{50}C_3 \] So the expression becomes: \[ ^{50}C_4 + ^{55}C_3 + ^{54}C_3 + ^{53}C_3 + ^{52}C_3 + ^{51}C_3 + ^{50}C_3 \] We will use Pascal's identity: \( ^nC_r + ^nC_{r-1} = ^{n+1}C_r \). (This can also be written as \( ^nC_r + ^nC_{r+1} = ^{n+1}C_{r+1} \)).
Rearrange the terms to apply the identity: \[ = (^{50}C_3 + ^{50}C_4) + ^{51}C_3 + ^{52}C_3 + ^{53}C_3 + ^{54}C_3 + ^{55}C_3 \] Apply Pascal's identity to the first two terms \( ^{50}C_3 + ^{50}C_4 \):
\( ^{50}C_3 + ^{50}C_4 = ^{50+1}C_4 = ^{51}C_4 \)
So the expression becomes: \[ = ^{51}C_4 + ^{51}C_3 + ^{52}C_3 + ^{53}C_3 + ^{54}C_3 + ^{55}C_3 \] Apply Pascal's identity again to \( ^{51}C_4 + ^{51}C_3 \):
\( ^{51}C_4 + ^{51}C_3 = ^{51+1}C_4 = ^{52}C_4 \)
The expression is now: \[ = ^{52}C_4 + ^{52}C_3 + ^{53}C_3 + ^{54}C_3 + ^{55}C_3 \] Continue this process:
\( (^{52}C_4 + ^{52}C_3) + ^{53}C_3 + ^{54}C_3 + ^{55}C_3 = ^{53}C_4 + ^{53}C_3 + ^{54}C_3 + ^{55}C_3 \)
\( (^{53}C_4 + ^{53}C_3) + ^{54}C_3 + ^{55}C_3 = ^{54}C_4 + ^{54}C_3 + ^{55}C_3 \)
\( (^{54}C_4 + ^{54}C_3) + ^{55}C_3 = ^{55}C_4 + ^{55}C_3 \)
Finally, apply Pascal's identity one last time:
\( ^{55}C_4 + ^{55}C_3 = ^{55+1}C_4 = ^{56}C_4 \)
In simple words: This problem asks us to add up several combination numbers. We used a special rule called Pascal's Identity many times to simplify the sum step by step. Each time, we combined two numbers into one, making the sum smaller until we reached the final answer.

🎯 Exam Tip: This type of problem is a classic application of Pascal's Identity (or Hockey-stick identity). When you see a sum of combinations where the upper index is increasing and the lower index is fixed, or vice versa, think about applying this identity repeatedly.

Question 11. \( ^{n-1}C_3 + ^{n-1}C_4 > ^nC_3 \) if
(a) \( n > 7 \)
(b) \( n \geq 7 \)
(c) \( n > 6 \)
(d) \( n \geq 6 \)
Answer: (a) \( n > 7 \)
Given the inequality: \( ^{n-1}C_3 + ^{n-1}C_4 > ^nC_3 \).
Using Pascal's Identity: \( ^kC_r + ^kC_{r+1} = ^{k+1}C_{r+1} \).
Here, for the left side, \( k = n-1 \), \( r = 3 \), and \( r+1 = 4 \).
So, \( ^{n-1}C_3 + ^{n-1}C_4 = ^{(n-1)+1}C_{4} = ^nC_4 \).
The inequality becomes:
\( ^nC_4 > ^nC_3 \)
We know that \( ^nC_r > ^nC_s \) if \( r > s \) and \( n > r+s \). A simpler way is to use the property: \( ^nC_r \) increases as \( r \) increases from \( 0 \) to \( n/2 \).
For \( ^nC_4 > ^nC_3 \), this means we are on the increasing side of the combinations for \( n \).
This implies that \( 4 > 3 \) (which is true), and also that \( 4 < n/2 \).
So, \( 4 < n/2 \)
Multiply both sides by 2:
\( 8 < n \)
\( n > 8 \)
However, the standard condition for \( ^nC_r > ^nC_{r-1} \) is \( n > 2r - 1 \).
Here, \( r=4 \), so \( n > 2(4) - 1 \)
\( n > 8 - 1 \)
\( n > 7 \)

Alternatively, let's use the explicit formula for combinations:
\( \frac{n!}{4!(n-4)!} > \frac{n!}{3!(n-3)!} \)
Since \( n! \) is positive, we can divide both sides by \( n! \):
\( \frac{1}{4!(n-4)!} > \frac{1}{3!(n-3)!} \)
Cross-multiply by positive terms:
\( 3!(n-3)! > 4!(n-4)! \)
We know that \( 4! = 4 \times 3! \) and \( (n-3)! = (n-3) \times (n-4)! \).
Substitute these:
\( 3! (n-3)(n-4)! > (4 \times 3!) (n-4)! \)
Divide both sides by \( 3! \) and \( (n-4)! \) (which are positive):
\( n-3 > 4 \)
Add 3 to both sides:
\( n > 7 \)
For the combinations to be defined, we must have \( n-1 \ge 4 \) (from \( ^{n-1}C_4 \)) and \( n \ge 3 \) (from \( ^nC_3 \)).
\( n-1 \ge 4 \implies n \ge 5 \).
So, combining \( n > 7 \) with \( n \ge 5 \), the condition \( n > 7 \) is the stricter one.
In simple words: First, we used a combination rule to make the left side of the inequality simpler. Then, we solved the new inequality using the definition of combinations or a property that compares combination values. This showed that 'n' must be larger than 7.

🎯 Exam Tip: When dealing with inequalities involving combinations, using Pascal's Identity to simplify expressions is a good first step. Then, either use the property \( ^nC_r > ^nC_{r-1} \implies n > 2r-1 \) or expand the factorial terms and simplify the inequality algebraically. Always consider the conditions for combinations to be defined (\( n \ge r \)).