If frac{1}{{ }^{5}C_{r}} + frac{1}{{ }^{6}C_{ | vedclass

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(B) Given, $\frac{1}{{ }^{5}C_{r}} + \frac{1}{{ }^{6}C_{r}} = \frac{1}{{ }^{4}C_{r}}$Using the formula ${ }^{n}C_{r} = \frac{n!}{r!(n-r)!}$, we have:$

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$2$

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(B) Given, $\frac{1}{{ }^{5}C_{r}} + \frac{1}{{ }^{6}C_{r}} = \frac{1}{{ }^{4}C_{r}}$Using the formula ${ }^{n}C_{r} = \frac{n!}{r!(n-r)!}$, we have:$\frac{r!(5-r)!}{5!} + \frac{r!(6-r)!}{6!} = \frac{r!(4-r)!}{4!}$Dividing by $r!$ and multiplying by $4!$:$\frac{(5-r)!}{5} + \frac{(6-r)(5-r)!}{6 	imes 5} = (4-r)!$Dividing by $(4-r)!$:$\frac{(5-r)}{5} + \frac{(6-r)(5-r)}{30} = 1$Multiply by $30$:$6(5-r) + (6-r)(5-r) = 30$$30 - 6r + 30 - 11r + r^{2} = 30$$r^{2} - 17r + 30 = 0$$(r-2)(r-15) = 0$Since $r \leq 4$ (as ${ }^{4}C_{r}$ is defined), we have $r = 2$.

If $\frac{1}{{ }^{5}C_{r}} + \frac{1}{{ }^{6}C_{r}} = \frac{1}{{ }^{4}C_{r}},$ then the value of $r$ is

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