If 4 red balls and 5 green balls are selected | vedclass

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Question

(A) The number of ways to select $4$ red balls from $n$ balls is ${}^{n}C_4$. The number of ways to select $5$ green balls from $n$ balls is ${}^{n}C_

Vedclass · Accepted Answer

$n > 8$

Vedclass · Answer

(A) The number of ways to select $4$ red balls from $n$ balls is ${}^{n}C_4$. The number of ways to select $5$ green balls from $n$ balls is ${}^{n}C_5$. The sum of both selections is ${}^{n}C_4 + {}^{n}C_5$. Using the Pascal's identity,${}^{n}C_r + {}^{n}C_{r-1} = {}^{n+1}C_r$,we have ${}^{n}C_4 + {}^{n}C_5 = {}^{n+1}C_5$. We are given that the sum is greater than ${}^{n+1}C_4$,so ${}^{n+1}C_5 > {}^{n+1}C_4$. Using the formula ${}^{m}C_r = \frac{m!}{r!(m-r)!}$,we get $\frac{(n+1)!}{5!(n-4)!} > \frac{(n+1)!}{4!(n-3)!}$. Simplifying this,we get $\frac{1}{5} > \frac{1}{n-3}$. This implies $n-3 > 5$,so $n > 8$.

If $4$ red balls and $5$ green balls are selected from $n$ balls,and the sum of both selections is greater than ${}^{n+1}C_4$,then the value of $n$ is equal to:

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