Question

nC(n-2)=36 then evaluate nC(n-5).​

Answer 1

Answer:

We know the property,

nCr = nCn-r

so, since

nCn-2=36

Or, nC2=36

Oh solving, we get n=9

So, nCn-5 = 9C4 = 9*8*7*6/24 = 126

Hope this helps you !

Answer 2

Answer:

The value of $n_{c_{n-5} }$ = 126.

Step-by-step explanation:

Given ncₙ₋₂ =36

We have to find the value of $n_{c_{n-5} }$ .

We know that $n_{c_{r} } =\frac{n!}{(n-r)!(r)!}$

Now $n_{c_{n-2} } = \frac{n!}{(n-n+2)!(n-2)!}$ = 36

⇒$\frac{n(n-1)(n-2)!}{2!(n-2)!} =36$

⇒$\frac{n(n-1)}{2} =36$

⇒n(n-1) = 36×2

⇒ n(n-1) = 72

⇒n(n-1) = 9 × 8

∴ n = 9

Now we will find the value of nc(n-5)

$n_{c_{n-5} }$

n=9

n-5 = 9-5=4

$9_{c_{4} } = \frac{9!}{(9-4)!(4)!} =\frac{9!}{5!(4)!} =\frac{(9)(8)(7)(6)5!}{5!4!} =\frac{(9)(8)(7)(6)}{(4)(3)(2)(1)}$ = 9×7×2 = 126

Therefore,  $n_{c_{n-5} }$ = 126.