If nCr – 1 = 36, nCr = 84, and nCr + 1 = 126, then find rC2. [Hint: Form equation using C(n, r-1), C(n, r), and C(n, r+1) to find the value of r.]

This is a short answer type question as classified in NCERT Exemplar G i v e n t h a t C r − 1 n = 3 6 … ( i ) C r n = 8 4 … ( i i ) C r + 1 n = 1 2 6 … ( i i i ) D i v i d i n g e q n . ( i ) b y e q n . ( i i ) w e g e t C r − 1 n C r n = 3 6 8 4 ⇒ n ! ( r − 1 ) ! ( n − r + 1 ) ! n ! r ! ( n − r ) ! = 3 7 [ ? C r n = n ! r ! ( n − r ) ! ] ⇒ n ! ( r − 1 ) ! ( n − r + 1 ) ! * r ! ( n − r ) ! n ! = 3 7 ⇒ r . ( r − 1 ) ! ( n − r ) ! ( r − 1 ) ! ( n − r + 1 ) ! + ( n − r ) ! = 3 7 ⇒ r n − r + 1 = 3 7 ⇒ 3 n − 3 r + 3 = 7 r ⇒ 3 n − 1 0 r = − 3 … ( i v ) D i v i d i n g e q n . ( i i ) b y e q n . ( i i i ) w e g e t C r n C r + 1 n = 8 4 1 2 6 ⇒ n ! r ! ( n − r ) ! n ! ( r + 1 ) ! ( n − r − 1 ) ! = 2 3 ⇒ &th

If ⁿC_{r – 1} = 36, ⁿC_r = 84, and ⁿC_{r + 1}= 126, then find ^rC₂.

[Hint: Form equation using C(n, r-1), C(n, r), and C(n, r+1) to find the value of r.]

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Payal Gupta | Contributor-Level 10

7 months ago

This is a short answer type question as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t {}_{n}{C_{r - 1}} = 36 \dots (i) \\ {}_{n}{C_{r}} = 84 \dots (i i) \\ {}_{n}{C_{r + 1}} = 126 \dots (i i i) \\ D i v i d i n g e q n . (i) b y e q n . (i i) w e g e t \\ \frac{{}_{n}{C_{r - 1}}}{{}_{n}{C_{r}}} = \frac{36}{84} \\ \Rightarrow \frac{\frac{n!}{(r - 1)! (n - r + 1)!}}{\frac{n!}{r! (n - r)!}} = \frac{3}{7} [? {}_{n}{C_{r}} = \frac{n!}{r! (n - r)!}] \\ \Rightarrow \frac{n!}{(r - 1)! (n - r + 1)!} * \frac{r! (n - r)!}{n!} = \frac{3}{7} \\ \Rightarrow \frac{r . (r - 1)! (n - r)!}{(r - 1)! (n - r + 1)! + (n - r)!} = \frac{3}{7} \\ \Rightarrow \frac{r}{n - r + 1} = \frac{3}{7} \Rightarrow 3 n - 3 r + 3 = 7 r \\ \Rightarrow 3 n - 10 r = - 3 \dots (i v) \\ D i v i d i n g e q n . (i i) b y e q n . (i i i) w e g e t \\ \frac{{}_{n}{C_{r}}}{{}_{n}{C_{r + 1}}} = \frac{84}{126} \Rightarrow \frac{\frac{n!}{r! (n - r)!}}{\frac{n!}{(r + 1)! (n - r - 1)!}} = \frac{2}{3} \\ \Rightarrow &th \end{array}$

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