Mathematics - Permutation & Combination Question with Solution | TestHub

Let $n$ be the number of sides.

Therefore, the number of diagonals is ${}^n{C}_2-n=54$ (given).

$$\frac{n(n-1)}{2}-n=54$$

$$n^2-n-2n=108$$

$$n^2-3n-108=0$$

$$(n-12)(n+9)=0$$

$$n=12(sincencannotbenegative)$$

(B) The number of divisors of the form $4n+2$ ($n\ge 0$) of the integer 240 is 0, 6, 10, 30. These are 4 in number.

Therefore, $B-p,B-r$.

(C)

$$63={}^{2n+1}{C}_1+{}^{2n+1}{C}_2+\dots +{}^{2n+1}{C}_n$$

Therefore,

$$64={}^{2n+1}{C}_0+{}^{2n+1}{C}_1+\dots +{}^{2n+1}{C}_n$$

$$=\frac{1}{2}\left({}^{2n+1}{C}_0+{}^{2n+1}{C}_1+\dots +{}^{2n+1}{C}_{2n+1}\right)=\frac{1}{2}\cdot 2^{2n+1}$$

$$128=2^{2n+1}$$

$$2^7=2^{2n+1}$$

$$7=2n+1$$

$$n=3$$

Therefore, $n+1=3+1=4$.

Therefore, $c-p,c-r$.