If n and r are integers such that 1-r-n- then n- C-n-1- r-1- - - Maths Q-A

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Inequality

If n and r ar...

Question

If $n$ and $r$ are integers such that $1 \leq r \leq n,$ then ${}_{n}{C^{(n - 1, r - 1)}} =$

$C^{(n, r)}$

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${}_{r}{C^{(n, r)}}$

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${}_{n}C^{(n, r)}$

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none of these

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Solution

The correct option is B
${}_{r}{C^{(n, r)}}$

Step: Explanation for correct option.
Option B:- ${}_{r}{C^{(n, r)}} .$
Solve for value of ${}_{n}{C^{(n - 1, r - 1)}}$
We know that
$C^{(n, r)} = \frac{n!}{r! (n - r)!}$
$C^{(n - 1, r - 1)} = \frac{(n - 1)!}{(r - 1)! (n - r)!}$
${}_{n}{C^{(n - 1, r - 1)}} = n \cdot \frac{(n - 1)!}{(r - 1)! (n - r)!}$
$= \frac{n!}{(r - 1)! (n - r)!}$ [Since $n \cdot (n - 1)! = n!$ ]
$= \frac{r \cdot n!}{r \cdot (r - 1)! (n - r)!}$ [Multiply Numerator and Denominator with $r$ ]
$= \frac{r \cdot n!}{r! (n - r)!} = r \cdot C (n, r)$
Hence, option B is the correct answer i.e., ${}_{r}{C^{(n, r)}} .$

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Similar questions

Q. Let r and n be positive integers such that 1 ≤ r ≤ n. Then prove the following:
(a)

\frac{{}^{n}C_{r}}{{}^{n}C_{r - 1}} = \frac{n - r + 1}{r}

(b) n ·_{^{n −}}_¹C_r_{− 1} = (n − r + 1) ⁿC_r_{− 1}
(c)

\frac{{}^{n}C_{r}}{{}^{n - 1}C_{r - 1}} = \frac{n}{r}

(iv) ⁿC_r + 2 · ⁿC_r_{− 1} + ⁿC_r_{− 2} = ^{n +}²C_r.

Let r and n be positive integers such that $1 \leq r \leq n .$ Then prove the following :

(i) $\frac{^{n} C_{r}}{^{n} C_{r - 1}} = \frac{n - r + 1}{r}$
(ii) $n^{n - 1} C_{r - 1} = (n - r + 1)^{n} C_{r - 1}$
(iii) $\frac{^{n} C_{r}}{^{n - 1} C_{r - 1}} = \frac{n}{r}$
(iv) $^{n} C_{r} + 2^{n} C_{r - 1} +^{n} C_{r - 2} =^{n + 2} C_{r}$

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If nand r are integers such that 1≤r≤n,then C(n-1,r-1)n=

If $n$ and $r$ are integers such that $1 \leq r \leq n,$ then ${}_{n}{C^{(n - 1, r - 1)}} =$