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Question Number 6090 by sanusihammed last updated on 13/Jun/16

Answered by Rasheed Soomro last updated on 13/Jun/16

$${}^{x+\frac{y}{2}}{\mathrm{C}}_2=21\wedge {}^{\frac{x+y}{2}}{\mathrm{P}}_2=30$$$$Letx+\frac{y}{2}=n,{}^n{\mathrm{C}}_2=21$$$$\frac{n!}{2!(n-2)!}=21$$$$\frac{n(n-1)}{1\times 2}=21$$$$n^2-n=42$$$$n^2-n-42=0$$$$(n-7)(n+6)=0$$$$n=7\mid n=-6\left(discardable\right)$$$$x+\frac{y}{2}=7\Rightarrow 2x+y=14\left(i\right)$$$$-----$$$$Suppose\frac{x+y}{2}=m$$$${}^{\frac{x+y}{2}}{\mathrm{P}}_2=30{\Rightarrow }^m{\mathrm{P}}_2=30$$$$\frac{m!}{(m-2)!}=30$$$$m(m-1)=30$$$$m^2-m-30=0$$$$(m-6)(m+5)=0$$$$m=6\mid m=-5\left(dicardable\right)$$$$\frac{x+y}{2}=6\Rightarrow x+y=12\left(ii\right)$$$$From\left(i\right)\mathrm{\&}\left(ii\right):$$$$x=2,y=10$$

Commented by sanusihammed last updated on 13/Jun/16

$$Thanksverymuch$$

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