If { }^{n+4} C_{n+1}-{ }^{n+3} C_n=15(n+2),th | vedclass

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(D) Given the equation: ${ }^{n+4} C_{n+1}-{ }^{n+3} C_n=15(n+2)$.Using the property ${ }^n C_r = { }^n C_{n-r}$,we have ${ }^{n+4} C_{n+1} = { }^{n+4

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$27$

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(D) Given the equation: ${ }^{n+4} C_{n+1}-{ }^{n+3} C_n=15(n+2)$.Using the property ${ }^n C_r = { }^n C_{n-r}$,we have ${ }^{n+4} C_{n+1} = { }^{n+4} C_{(n+4)-(n+1)} = { }^{n+4} C_3$ and ${ }^{n+3} C_n = { }^{n+3} C_{(n+3)-n} = { }^{n+3} C_3$.Expanding the combinations: $\frac{(n+4)(n+3)(n+2)}{3 	imes 2 	imes 1} - \frac{(n+3)(n+2)(n+1)}{3 	imes 2 	imes 1} = 15(n+2)$.Dividing both sides by $(n+2)$ (since $n+2 
eq 0$): $\frac{(n+4)(n+3)}{6} - \frac{(n+3)(n+1)}{6} = 15$.Multiply by $6$: $(n^2+7n+12) - (n^2+4n+3) = 90$.$3n + 9 = 90$.$3n = 81$.$n = 27$.

If ${ }^{n+4} C_{n+1}-{ }^{n+3} C_n=15(n+2)$,then $n=$

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