Let 0 leq r leq n. If { }^{n+1} C_{r+1} : { } | vedclass

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(C) Given the ratio ${ }^{n+1} C_{r+1} : { }^{n} C_{r} : { }^{n-1} C_{r-1} = 55 : 35 : 21$.First,consider the ratio $\frac{{ }^{n+1} C_{r+1}}{{ }^{n}

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$50$

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(C) Given the ratio ${ }^{n+1} C_{r+1} : { }^{n} C_{r} : { }^{n-1} C_{r-1} = 55 : 35 : 21$.First,consider the ratio $\frac{{ }^{n+1} C_{r+1}}{{ }^{n} C_{r}} = \frac{55}{35} = \frac{11}{7}$.Using the formula ${ }^{n} C_{r} = \frac{n!}{r!(n-r)!}$,we get:$\frac{(n+1)!}{(r+1)!(n-r)!} 	imes \frac{r!(n-r)!}{n!} = \frac{n+1}{r+1} = \frac{11}{7}$.$7n + 7 = 11r + 11 \implies 7n - 11r = 4$  $(1)$.Next,consider the ratio $\frac{{ }^{n} C_{r}}{{ }^{n-1} C_{r-1}} = \frac{35}{21} = \frac{5}{3}$.$\frac{n!}{r!(n-r)!} 	imes \frac{(r-1)!(n-r)!}{(n-1)!} = \frac{n}{r} = \frac{5}{3}$.$3n = 5r \implies n = \frac{5r}{3}$  $(2)$.Substitute $(2)$ into $(1)$:$7(\frac{5r}{3}) - 11r = 4$.$\frac{35r - 33r}{3} = 4 \implies 2r = 12 \implies r = 6$.Then $n = \frac{5(6)}{3} = 10$.Finally,calculate $2n + 5r = 2(10) + 5(6) = 20 + 30 = 50$.

Let $0 \leq r \leq n$. If ${ }^{n+1} C_{r+1} : { }^{n} C_{r} : { }^{n-1} C_{r-1} = 55 : 35 : 21$,then $2n + 5r$ is equal to:

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