Find the value of $^{n} C_{n - r}$ .

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B
$^{n} C_{r} = \frac{n!}{r! (n - r)!}, 0 \leq r \leq n$
$^{n} C_{n - r} = \frac{n!}{(n - r)! (n - (n - r))!} = \frac{n!}{(n - r)! r!} =^{n} C_{r}$
$^{n} C_{r} =^{n} C_{n - r}$

Suggest Corrections

Similar questions

Q. The value of

^{n} C_{n}

Q. Given that

^{n} C_{n - r} + 3^{n} C_{n - r + 1} + 3.^{n} C_{n - r + 2} +^{n} C_{n - r + 3} =^{x} C_{r}

. Let

x = n + k

, then find

k

Find the coefficient of $x^{n - 2}$ in
$(^{n} C_{0} +^{n} C_{1} x +^{n} C_{2} x^{2} . . . . .^{n} C_{n} x^{n}) \times (^{n} C_{0} +^{n} C_{1} x +^{n} C_{2} x^{2} . . . . .^{n} C_{n} x^{n})$

Q. If

^{n} C_{n - r} + 3^{n} C_{n - r + 1} + 3^{n} C_{n - r + 2} +^{n} C_{n - r + 3} =^{p} C_{r}

then prove that

p = n + 3

here r/nCr=(n-r)/nCn-r how it is possble because its true that nCr=nCn-r but how n-r=r as in numerator

Join BYJU'S Learning Program

Find the value of nCn−r.

The correct option is B nCr=n!r!(n−r)!,0≤r≤n nCn−r=n!(n−r)!(n−(n−r))!=n!(n−r)!r!=nCr nCr=nCn−r

Find the value of $^{n} C_{n - r}$ .

The correct option is B
$^{n} C_{r} = \frac{n!}{r! (n - r)!}, 0 \leq r \leq n$
$^{n} C_{n - r} = \frac{n!}{(n - r)! (n - (n - r))!} = \frac{n!}{(n - r)! r!} =^{n} C_{r}$
$^{n} C_{r} =^{n} C_{n - r}$