If 1 - r - n then n n -1 C r -1 is equal to -A- n-rn Cr-1B- n-1n Cr-1C- n-r-1n Cr-1D- n-r-1n Cr-1

The correct option is C n^n 1C_r 1=n(n 1)!/n 1 (r 1)!(r 1)! n^n 1C_r 1=(n)!/(n 1)!(r 1)! n^n 1C_r 1=(n r+1)n!/(n r+1)(n r)!(r 1)! Since, all the options havi ...

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Byju's Answer

Standard XI

Mathematics

Binomial Coefficients

If 1 ≤ r ≤ n ...

Question

If $1 \leq r \leq n$ then $n^{n - 1} C_{r - 1}$ is equal to .........

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Solution

The correct option is D

$n^{n} - 1 C_{r - 1} = n \frac{(n - 1)!}{n - 1 - (r - 1)! (r - 1)!}$
$n^{n} - 1 C_{r - 1} = \frac{(n)!}{(n - 1)! (r - 1)!}$
$n^{n} - 1 C_{r - 1} = \frac{(n - r + 1) n!}{(n - r + 1) (n - r)! (r - 1)!}$

Since, all the options having $^{n} C_{r - 1}$ ,
Simplify the expression to get $^{n} C_{r - 1}$ in the expression
$n^{n - 1} C_{r - 1} = (n - r + 1) {\frac{n!}{(n - r + 1)! (r - 1)!}}$
$n^{n - 1} C_{r - 1} = (n - r + 1) {\frac{n!}{{n - (r - 1)}! (r - 1)!}}$
$n^{n - 1} C_{r - 1} = (n - r + 1)^{n} C_{r - 1}$

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Similar questions

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) $^{n} C_{r} + 2^{n} C_{r - 1} +^{n} C_{r - 2} =^{n + 2} C_{r}$