Question

If nC8 = nC3 find nC2
​

Answer 1

Answer:

simple And Answer is

Step-by-step explanation:

w.k.t nCr=nCn-r

so nC8=nCn-3

it will become 8=n-3

n=11

So By Combination

11C2 = 11!/2!(11-2)!

= 11*10*9!/2!*9! (9! & 9! WILL GET CANCEL)

So now we have 11*10/2

And Hence the Answer Is 22.

Thank you

Answer 2

Answer:

The value of $n$ is $11$ and the value of $n_C_2$, i.e., $11_C_2$ is $55$.

Step-by-step explanation:

We know that $n_{C}_r = \frac{n!}{(n-r)!r!}$ ....... (1)

Also, $n_{C}_{n-r} = \frac{n!}{(n-(n-r))!(n-r)!}$

⇒ $n_{C}_{n-r} = \frac{n!}{r!(n-r)!}$ ...... (2)

Observe that equation (1) and (2) are equal.

This implies, $n_{C}_{n-r} = n_{C}_r$ ...... (3)

Consider the hypothesis as follows:

$n_{C}_8 = n_{C}_3$

Rewrite the condition as follows:

⇒ $n_{C}_8 = n_{C}_{n-3}$ (From 3)

⇒ $8=n-3$

⇒ $n=11$

Thus, the value of $n$ is $11$.

Now, find the value of $11_C_2$.

$11_C_2=\frac{11!}{(11-2)!2!}$

       $=\frac{11!}{9!2!}$

Simplify as follows:

$11_C_2=\frac{11 \cdot 10 \cdot9!}{9!2 \cdot1}$

        $=\frac{11 \cdot10}{2}$

        $=55$

Therefore, the value of $n_C_2$, i.e., $11_C_2$ is $55$.

#SPJ3