Class 11: Combinations – Exercise 17.1 – ICSE / ISC / CBSE Mathematics Portal for K12 Students

Question 1: Evaluate the following:

$\displaystyle \text{i) } ^{14} \rm C_3$ $\displaystyle \text{ii) } ^{12} \rm C_{10}$ $\displaystyle \text{iii) } ^{35} \rm C_{35}$ $\displaystyle \text{iv) } ^{n+1} \rm C_n$

$\displaystyle \text{v) } \sum \limits_{r=1}^{5} \ \displaystyle ^{5} \rm C_{r}$

Answer:

$\displaystyle \text{i) } ^{14} \rm C_3 = \frac{14!}{3! (14-3)!} = \frac{14!}{3! 11!} = \frac{14 \times 13 \times 12}{6} = 364$

$\displaystyle \text{ii) } ^{12} \rm C_{10} = \frac{12!}{10! (12-10)!} = \frac{12!}{10! 2!} = \frac{11 \times 11 }{2} = 66$

$\displaystyle \text{iii) } ^{35} \rm C_{35} = \frac{35!}{35! (35-35)!} = \frac{35!}{35! 0!} = 1$

$\displaystyle \text{iv) } ^{n+1} \rm C_n = \frac{(n+1)!}{n! (n+1-n)!} = \frac{(n+1)n!}{n!} = (n+1)$

$\displaystyle \text{v) } \sum \limits_{r=1}^{5} \ \displaystyle ^{5} \rm C_{r}$

$= ^{5} \rm C_{1} + ^{5} \rm C_{2} + ^{5} \rm C_{3}+^{5} \rm C_{4}+^{5} \rm C_{5}$

$\displaystyle = \frac{5!}{5! (5-1)!} + \frac{5!}{5! (5-2)!} + \frac{5!}{5! (5-3)!} + \frac{5!}{5! (5-4)!} + \frac{5!}{5! (5-5)!}$

$\displaystyle = 5 + 10 + 10 + 5 + 1 = 31$

$\displaystyle \\$

Question 2: If $\displaystyle ^{n} \rm C_{12} = ^{n} \rm C_{5}$ , find the value of $\displaystyle n$ .

Answer:

$\displaystyle \text{Given } ^{n} \rm C_{12} = ^{n} \rm C_{5}$

Applying the formula, when $\displaystyle ^{n} \rm C_{p} = ^{n} \rm C_{q} \Rightarrow p+q = n$

Therefore $\displaystyle 12 + 5 = n \Rightarrow n = 17$

$\displaystyle \\$

Question 3: If $\displaystyle ^{n} \rm C_{4} = ^{n} \rm C_{6}$ , find $\displaystyle ^{12} \rm C_{n}$ .

Answer:

$\displaystyle \text{Given } ^{n} \rm C_{4} = ^{n} \rm C_{6}$

Applying the formula, when $\displaystyle ^{n} \rm C_{p} = ^{n} \rm C_{q} \Rightarrow p+q = n$

Therefore $\displaystyle 4 + 6 = n \Rightarrow n = 10$

$\displaystyle ^{12} \rm C_{10} = \frac{12!}{10! 2!} = \frac{12 \times 11}{2} = 66$

$\displaystyle \\$

Question 4: If $\displaystyle ^{n} \rm C_{10} = ^{n} \rm C_{12}$ , find $\displaystyle ^{23} \rm C_{n}$ .

Answer:

$\displaystyle \text{Given } ^{n} \rm C_{10} = ^{n} \rm C_{12}$

Applying the formula, when $\displaystyle ^{n} \rm C_{p} = ^{n} \rm C_{q} \Rightarrow p+q = n$

Therefore $\displaystyle 10 + 12 = n \Rightarrow n = 22$

$\displaystyle ^{23} \rm C_{22} = \frac{23!}{22! (23-22)!} = \frac{23! }{22!} = 23$

$\displaystyle \\$

Question 5: If $\displaystyle ^{24} \rm C_{x} = ^{24} \rm C_{2x+3}$ , find $\displaystyle x$ .

Answer:

$\displaystyle \text{Given } ^{24} \rm C_{x} = ^{24} \rm C_{2x+3}$

Applying the formula, when $\displaystyle ^{n} \rm C_{p} = ^{n} \rm C_{q} \Rightarrow p+q = n$

Therefore $\displaystyle x + 2x+3 = 24 \Rightarrow x = 7$

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Question 6: If $\displaystyle ^{18} \rm C_{x} = ^{18} \rm C_{x+2}$ , find $\displaystyle x$ .

Answer:

$\displaystyle \text{Given } ^{18} \rm C_{x} = ^{18} \rm C_{x+2}$

Applying the formula, when $\displaystyle ^{n} \rm C_{p} = ^{n} \rm C_{q} \Rightarrow p+q = n$

Therefore $\displaystyle x + x+2 = 18 \Rightarrow x = 8$

$\displaystyle \\$

Question 7: If $\displaystyle ^{15} \rm C_{3r} = ^{15} \rm C_{r+3}$ , find $\displaystyle r$ .

Answer:

Given $\displaystyle ^{15} \rm C_{3r} = ^{15} \rm C_{r+3}$

Applying the formula, when $\displaystyle ^{n} \rm C_{p} = ^{n} \rm C_{q} \Rightarrow p+q = n$

Therefore $\displaystyle 3r+r+3 = 15 \Rightarrow r = 3$

$\displaystyle \\$

Question 8: If $\displaystyle ^{8} \rm C_{r} - ^{7} \rm C_{3} = ^{7} \rm C_{2}$ , find $\displaystyle r$ .

Answer:

$\displaystyle ^{8} \rm C_{r} - ^{7} \rm C_{3} = ^{7} \rm C_{2}$

$\displaystyle \Rightarrow ^{8} \rm C_{r} = ^{7} \rm C_{2} + ^{7} \rm C_{3}$

$\displaystyle \Rightarrow \frac{8!}{r! (8-r)!} = \frac{7!}{2! 5!} + \frac{7!}{3! 4!}$

$\displaystyle \Rightarrow \frac{8!}{r! (8-r)!} = \frac{7 \times 6}{2} + \frac{7 \times 6 \times 5}{6}$

$\displaystyle \Rightarrow \frac{8!}{r! (8-r)!} = 56$

$\displaystyle \Rightarrow r! (8-r)! = 720$

$\displaystyle \Rightarrow r! (8-r)! = 3! \times 5!$

Comparing LHS with RHS we get,

If $\displaystyle r = 3$ and $\displaystyle 8-r = 5 \Rightarrow r = 3$

If $\displaystyle r = 5$ , then $\displaystyle 8-r = 3 \Rightarrow r = 5$

Hence $\displaystyle r$ can be $\displaystyle 3$ or $\displaystyle 5$ .

$\displaystyle \\$

Question 9: If $\displaystyle ^{15} \rm C_{r} : ^{15} \rm C_{r-1} = 11:5$ , find $\displaystyle r$ .

Answer:

$\displaystyle ^{15} \rm C_{r} : ^{15} \rm C_{r-1} = 11:5$

$\displaystyle \Rightarrow \frac{15!}{r! (15-r)!} \times \frac{(r-1)! ( 15-r+1)!}{15!} = \frac{11}{5}$

$\displaystyle \Rightarrow \frac{(r-1)! ( 16-r) (15-r)!}{r(r-1)!(15-r)!} = \frac{11}{5}$

$\displaystyle \Rightarrow \frac{16-r}{r} = \frac{11}{5}$

$\displaystyle \Rightarrow 80-5r = 11r$

$\displaystyle \Rightarrow 80 = 16 r$

$\displaystyle \Rightarrow r = 5$

$\displaystyle \\$

Question 10: If $\displaystyle ^{n+2} \rm C_{8} : ^{n-2} \rm P_{4} = 57:16$ , find $\displaystyle r$ .

Answer:

$\displaystyle ^{n+2} \rm C_{8} : ^{n-2} \rm P_{4} = 57:16$

$\displaystyle \Rightarrow \frac{(n+2)!}{8! (n-6)!} \times \frac{(n-6)!}{(n-2)!} = \frac{57}{16}$

$\displaystyle \Rightarrow \frac{(n+2)(n+1)(n)(n-1)(n-2)!}{8! (n-2)!} = \frac{57}{16}$

$\displaystyle \Rightarrow (n+2)(n+1)(n)(n-1) = \frac{8!}{16} \times 57 = 7 \times 6 \times 5 \times 4 \times 3 \times 19 \times 3$

$\displaystyle = 18 \times 19 \times 20 \times 21$

$\displaystyle \Rightarrow n = 18$

$\displaystyle \\$

Question 11: If $\displaystyle ^{28} \rm C_{2r} : ^{24} \rm C_{2r-4} = 225:11$ , find $\displaystyle r$ .

Answer:

$\displaystyle ^{28} \rm C_{2r} : ^{24} \rm C_{2r-4} = 225:11$

$\displaystyle \Rightarrow \frac{28!}{(2r)! (28-r)!} \times \frac{(2r-4)! (28-2r)!}{24!} = \frac{225}{11}$

$\displaystyle \Rightarrow \frac{28 \times 27 \times 26 \times 25 \times ( 2r-4)!}{(2r) ( 2r-1)(2r-2)(2r-3)(2r-4)!} = \frac{225}{11}$

$\displaystyle \Rightarrow \frac{28 \times 27 \times 26 \times 25 }{(2r) ( 2r-1)(2r-2)(2r-3)} = \frac{225}{11}$

$\displaystyle \Rightarrow (2r) ( 2r-1)(2r-2)(2r-3) = 28 \times 33 \times 26 = 11 \times 12 \times 13 \times 14$

$\displaystyle \Rightarrow 2r-3 = 11 \Rightarrow r = 7$

$\displaystyle \\$

Question 12: If $\displaystyle ^{n} \rm C_{4}, \ \ ^{n} \rm C_{5}$ and $\displaystyle ^{n} \rm C_{6}$ are in AP, then find $\displaystyle n$ .

Answer:

$\displaystyle ^{n} \rm C_{4}, \ \ ^{n} \rm C_{5}$ and $\displaystyle ^{n} \rm C_{6}$ are in AP

$\displaystyle \Rightarrow ^{n} \rm C_{5} - ^{n} \rm C_{4} = ^{n} \rm C_{6} - ^{n} \rm C_{5}$

$\displaystyle \Rightarrow \frac{n!}{5! (n-5)!} - \frac{n!}{4! (n-4)!} = \frac{n!}{6! (n-6)!} - \frac{n!}{5! (n-5)!}$

$\displaystyle \Rightarrow \frac{1}{5! (n-5)!} - \frac{1}{4! (n-4)!} = \frac{1}{6! (n-6)!} - \frac{1}{5! (n-5)!}$

$\displaystyle \Rightarrow \frac{2}{5! (n-5)!} = \frac{1}{4! (n-4)!} + \frac{1}{6! (n-6)!}$

$\displaystyle \Rightarrow \frac{2}{5 \times 4! (n-5)(n-6)!} = \frac{1}{4! (n-4)(n-5)(n-6)!} + \frac{1}{6 \times 5 \times 4! (n-6)!}$

$\displaystyle \Rightarrow \frac{2}{5(n-5)} = \frac{1}{(n-4)(n-5)} + \frac{1}{30}$

$\displaystyle \Rightarrow \frac{1}{n-5} \Big[ \frac{2}{5} - \frac{1}{n-4} \Big] = \frac{1}{30}$

$\displaystyle \Rightarrow \frac{1}{n-5} \Big[ \frac{2n-8-5}{5(n-4)} \Big] = \frac{1}{30}$

$\displaystyle 30(2n-13) = 5 ( n-4)(n-5)$

$\displaystyle 12n - 78 = n^2 - 9n + 20$

$\displaystyle n^2 - 21n +98 = 0$

$\displaystyle (n-7)(n-14) = 0$

$\displaystyle n = 7$ or $\displaystyle n = 14$

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Question 13: If $\displaystyle ^{2n} \rm C_{3} : ^{n} \rm C_{2} = 44:3$ , find $\displaystyle n$ .

Answer:

Given $\displaystyle ^{2n} \rm C_{3} : ^{n} \rm C_{2} = 44:3$

$\displaystyle \Rightarrow \frac{(2n)!}{3! (2n-3)!} \times \frac{2! (n-2)!}{n!} = \frac{44}{3}$

$\displaystyle \Rightarrow \frac{(2n)(2n-1)(2n-2)}{3!} \times \frac{2!}{n(n-1)} = \frac{44}{3}$

$\displaystyle \Rightarrow \frac{(2n-1)(2n-2)}{2(n-1)} = 11$

$\displaystyle 2n-1 = 11 \Rightarrow n = 6$

$\displaystyle \\$

Question 14: If $\displaystyle ^{16} \rm C_{r} = ^{16} \rm C_{r+2}$ , find $\displaystyle ^{r} \rm C_{4}$ .

Answer:

Given $\displaystyle ^{16} \rm C_{r} = ^{16} \rm C_{r+2}$

Applying the formula, when $\displaystyle ^{n} \rm C_{p} = ^{n} \rm C_{q} \Rightarrow p+q = n$

Therefore $\displaystyle r + ( r + 2) = 16 \Rightarrow r = 7$

$\displaystyle ^{7} \rm C_{4} = \frac{7!}{4! (7-4)!} = \frac{7 \times 6 \times 5 }{6} = 35$

$\displaystyle \\$

Question 15: If $\displaystyle \alpha = ^{m} \rm C_{2}$ , then find the value of $\displaystyle ^{\alpha} \rm C_{2}$

Answer:

Given $\displaystyle \alpha = ^{m} \rm C_{2} \Rightarrow \alpha = \frac{m!}{2! (m-2)!} = \frac{m(m-1)}{2}$

$\displaystyle ^{\alpha} \rm C_{2} = \frac{\alpha !}{2! (\alpha - 2)!}$

$\displaystyle = \frac{\alpha ( \alpha - 1)}{2}$

$\displaystyle = \frac{1}{2} \times \frac{m(m-1)}{2} \times \Big[ \frac{m(m-1)}{2} - 1 \Big]$

$\displaystyle = \frac{m(m-1)}{4} \times \Big[ \frac{m^2 - m - 2}{2} \Big]$

$\displaystyle = \frac{1}{8} [ (m-1)(m^3 - m^2 - 2m) ]$

$\displaystyle = \frac{1}{8} [ m^4 - m^3 - 2m^2 - m^3 + m^2+ 2m]$

$\displaystyle = \frac{1}{2} [ m^4 - 2m^3 - m^2 + 2m ]$

$\displaystyle = \frac{1}{8} (m^2-2m)(m^2 -1)$

$\displaystyle = \frac{1}{8} m ( m-2) ( m-1)(m+1)$

$\displaystyle \\$

Question 16: Prove that the product of $\displaystyle 2n$ consecutive negative integers is divisible by $\displaystyle ( 2n)!$

Answer:

Let $\displaystyle 2n$ negative integers be $\displaystyle (-r), ( -r-1), ( -r-2), \ldots, ( - r - 2n + 1)$

Product of $\displaystyle 2n$ negative integers $\displaystyle = ( -1)^{2n} [ r ( r+1) (r+2) \ldots ( r+2n-1)]$

$\displaystyle = (-1)^{2n} \Big[ \frac{(r-1)! r ( r+1) (r+2) \ldots ( r+2n-1)}{(r-1)!} \Big]$

$\displaystyle = \frac{(r+2n-1)!}{(r-1)!} \times \frac{(2n)!}{(2n)!}$

$\displaystyle = ^{r+2n-1} \rm C_{2n} \cdot (2n)!$

Therefore product of $\displaystyle 2n$ consecutive negative integers is divisible by $\displaystyle ( 2n)!$

$\displaystyle \\$

Question 17: For all positive integers $\displaystyle n$ , show that

$\displaystyle ^{2n} \rm C_n + ^{2n} \rm C_{n-1} = \frac{1}{2} \big( ^{2n+2} \rm C_{n+1} \big)$

Answer:

$\displaystyle \text{LHS } = ^{2n} \rm C_n + ^{2n} \rm C_{n-1}$

$\displaystyle = \frac{(2n)!}{n! n!} + \frac{(2n)!}{(n-1)! (n+1)!}$

$\displaystyle = \frac{(2n)!}{(n-1)! n!} \Big[ \frac{1}{n} + \frac{1}{n+1} \Big]$

$\displaystyle = \frac{(2n)!}{(n-1)! n!} \Big[ \frac{2n+1}{n(n+1)} \Big]$

$\displaystyle = \frac{(2n+1)!}{(n+1)! n!}$

$\displaystyle \text{RHS } = \frac{1}{2} \Big[ ^{2n+2} \rm C_{n+1} \Big]$

$\displaystyle = \frac{1}{2} \Big[ \frac{(2n+2)!}{(n+1)! (n+1)!} \Big]$

$\displaystyle = \frac{1}{2} \Big[ \frac{(2n+2)(2n+1)!}{(n+1) n! (n+1)!} \Big]$

$\displaystyle = \frac{(2n+1)!}{(n+1)! n!}$

Therefore LHS = RHS hence proved.

$\displaystyle \\$

Question 18: Prove that: $\displaystyle ^{4n} \rm C_{2n} : ^{2n} \rm C_{n} = [ 1 \cdot 3 \cdot 5 \ldots (4n-1) ] : [ 1 \cdot 3 \cdot 5 \ldots (2n-1) ]^2$

Answer:

$\displaystyle \text{LHS } = ^{4n} \rm C_{2n} : ^{2n} \rm C_{n}$

$\displaystyle = \frac{^{4n} \rm C_{2n}}{^{2n} \rm C_{n}}$

$\displaystyle = \frac{(4n)!}{(2n)! (2n)!} \times \frac{(n)! (n)!}{(2n)!}$

$\displaystyle = \frac{(4n)(4n-1)(4n-2) \ldots (3) (2) (1)}{[ (2n)(2n-1)(2n-2) \ldots (3)(2)(1)]^2} \frac{(n!)^2}{(2n)!}$

$\displaystyle = \frac{(n!)^2}{(2n)!} \frac{ [ 1 \cdot 3 \cdot 5 \ldots (4n-3)(4n-1) ] [ 2 \cdot 4 \cdot 6 \ldots (4n-2)(4n) ] }{ [ 1 \cdot 3 \cdot 5 \ldots (2n-3)(2n-1) ]^2 [ 2 \cdot 4 \cdot 6 \ldots (2n-2)(2n) ]^2 }$

$\displaystyle = \frac{(n!)^2}{(2n)!} \frac{ [ 1 \cdot 3 \cdot 5 \ldots (4n-3)(4n-1) ] 2^{2n} [ 1 \cdot 2 \cdot 3 \ldots (2n-1)(2n) ] }{ [ 1 \cdot 3 \cdot 5 \ldots (2n-3)(2n-1) ]^2 2^{2n} [ 1 \cdot 2 \cdot 3 \ldots (n-1)(n) ]^2 }$

$\displaystyle = \frac{(n!)^2}{(2n)!} \frac{(2n)!}{(n!)^2} \frac{ [ 1 \cdot 3 \cdot 5 \ldots (4n-3)(4n-1) ] }{ [ 1 \cdot 3 \cdot 5 \ldots (2n-3)(2n-1) ]^2 }$

$\displaystyle = \frac{ [ 1 \cdot 3 \cdot 5 \ldots (4n-3)(4n-1) ] }{ [ 1 \cdot 3 \cdot 5 \ldots (2n-3)(2n-1) ]^2 } =$ RHS. Hence proved.

$\displaystyle \\$

$\displaystyle \text{Question 19: Evaluate: } ^{20} \rm C_{5} + \sum \limits_{r=2}^{5} \ \displaystyle ^{25-r} \rm C_{4}$

Answer:

$\displaystyle \text{LHS } = ^{20} \rm C_{5} + \sum \limits_{r=2}^{5}$

$\displaystyle = \ ^{20} \rm C_{5} + ^{23} \rm C_{4} +^{22} \rm C_{4} +^{21} \rm C_{4} + ^{20} \rm C_{4}$

$\displaystyle = \ \big( ^{20} \rm C_{4} + ^{20} \rm C_{5} \big) + \big( ^{21} \rm C_{4} +^{22} \rm C_{4} +^{23} \rm C_{4} \big)$

$\displaystyle \text{Since } ^{n} \rm C_{r-1} + ^{n} \rm C_{r} = ^{n+1} \rm C_{r}$

$\displaystyle = \ ^{21} \rm C_{5} + ^{21} \rm C_{4} +^{22} \rm C_{4} +^{23} \rm C_{4}$

$\displaystyle = \ ^{22} \rm C_{5} + ^{22} \rm C_{4} +^{23} \rm C_{4}$

$\displaystyle = \ ^{23} \rm C_{5} + ^{23} \rm C_{4}$

$\displaystyle = \ ^{24} \rm C_{5}$

$\displaystyle = \frac{24!}{5! 9!} = \frac{24 \times 23 \times 22 \times 21 \times 20}{5 \times 4 \times 3 \times 2 \times 1 } = 4 \times 23 \times 22 \times 21 = 41504$

$\displaystyle \\$

Question 20: Let $\displaystyle r$ and $\displaystyle n$ be positive integers such that $\displaystyle 1 \leq r \leq n$ . The prove the following:

$\displaystyle \text{i) } \frac{^{n} \rm C_{r}}{^{n} \rm C_{r-1}} = \frac{n-r+1}{r}$

$\displaystyle \text{ii) } n ^{n-1} \rm C_{r-1} = (n-r+1) ^{n} \rm C_{r-1}$

$\displaystyle \text{iii) } \frac{^{n} \rm C_{r}}{^{n-1} \rm C_{r-1}} = \frac{n}{r}$

$\displaystyle \text{iv) } ^{n} \rm C_{r} + 2 ^{n} \rm C_{r-1} + ^{n} \rm C_{r-2} = ^{n+2} \rm C_{r}$

Answer:

$\displaystyle \text{i) } \frac{^{n} \rm C_{r}}{^{n} \rm C_{r-1}} = \frac{n!}{r! (n-r)!} \times \frac{(r-1)! (n-r+1)!}{n!} \\ \\ = \frac{(r-1)!}{r! (r-1)!} \frac{(n-r+1)! (n-r)!}{(n-r)!} = \frac{n-r+1}{r}$