n C r - n C r -1-A- n-r-1-n-r-1

The correct option is A ^nC_r=n!/r!(n r)! ^nC_r 1=n!/(r 1)!(n r+1)! Substitute the value of ^nC_r and ^nC_r 1 ^nC_r/^nC_r 1=n!/r!(n r)!/n!/(r 1)!(n r+1)! =n!/ ...

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Byju's Answer

Standard XII

Mathematics

Higher Order Derivatives

n C r / n C r...

Question

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

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Solution

The correct option is A

$^{n} C_{r} = \frac{n!}{r! (n - r)!}$
$^{n} C_{r - 1} = \frac{n!}{(r - 1)! (n - r + 1)!}$
Substitute the value of $^{n} C_{r}$ and $^{n} C_{r - 1}$
$\frac{^{n} C_{r}}{^{n} C_{r - 1}} = \frac{\frac{n!}{r! (n - r)!}}{\frac{n!}{(r - 1)! (n - r + 1)!}}$
$= \frac{\frac{n!}{r \times (r - 1)! (n - r)!}}{\frac{n!}{(r - 1)! (n - r + 1) (n - r)!}}$
$= \frac{n - r + 1}{r}$

Suggest Corrections

Similar questions

$^{n} C_{r} \div^{n} C_{r - 1}$ =

$^{n} C_{r} \div^{n} C_{r - 1} =$

[MP PET 1984]

Q. Prove that :

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

Q. If

^{n} C_{r - 1} = 36,^{n} C_{r} = 84,^{n} C_{r + 1} = 126

, then

(n, r)

=

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nCrnCr−1=?

The correct option is A nCr=n!r!(n−r)! nCr−1=n!(r−1)!(n−r+1)! Substitute the value of nCr and nCr−1 nCrnCr−1=n!r!(n−r)!n!(r−1)!(n−r+1)! =n!r×(r−1)!(n−r)!n!(r−1)!(n−r+1)(n−r)! =n−r+1r

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