If for some m, n,{ }^6 C_{m}+2({ }^6 C_{m+1}) | vedclass

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(D) Given the inequality: ${ }^6 C_{m} + 2({ }^6 C_{m+1}) + { }^6 C_{m+2} > { }^8 C_3$. Using the identity ${ }^n C_r + { }^n C_{r+1} = { }^{n+1} C_{r

Vedclass · Accepted Answer

$372$

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(D) Given the inequality: ${ }^6 C_{m} + 2({ }^6 C_{m+1}) + { }^6 C_{m+2} > { }^8 C_3$. Using the identity ${ }^n C_r + { }^n C_{r+1} = { }^{n+1} C_{r+1}$,we get: $({ }^6 C_{m} + { }^6 C_{m+1}) + ({ }^6 C_{m+1} + { }^6 C_{m+2}) > { }^8 C_3$ ${ }^7 C_{m+1} + { }^7 C_{m+2} > { }^8 C_3$ ${ }^8 C_{m+2} > { }^8 C_3$. Since ${ }^8 C_3 = 56$,we need ${ }^8 C_{m+2} > 56$. For $m=2$,${ }^8 C_4 = 70 > 56$. Thus,$m=2$. Now,given the ratio: ${ }^{n-1} P_3 : { }^n P_4 = 1 : 8$. $\frac{(n-1)!}{(n-4)!} : \frac{n!}{(n-4)!} = 1 : 8$ $\frac{n-1}{n} = \frac{1}{8} \implies 8n - 8 = n \implies 7n = 8$ (Wait,re-evaluating: $\frac{(n-1)(n-2)(n-3)}{n(n-1)(n-2)(n-3)} = \frac{1}{n} = \frac{1}{8} \implies n=8$). Finally,calculate ${ }^n P_{m+1} + { }^{n+1} C_m = { }^8 P_3 + { }^9 C_2$. ${ }^8 P_3 = 8 	imes 7 	imes 6 = 336$. ${ }^9 C_2 = \frac{9 	imes 8}{2} = 36$. Sum $= 336 + 36 = 372$.

If for some $m, n$,${ }^6 C_{m}+2({ }^6 C_{m+1})+{ }^6 C_{m+2} > { }^8 C_3$ and ${ }^{n-1} P_3 : { }^n P_4 = 1 : 8$,then ${ }^n P_{m+1} + { }^{n+1} C_m$ is equal to

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