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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

$27$

Vedclass · Answer

(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 $n=$

Similar Questions

If $\frac{n!}{2!(n-2)!}$ and $\frac{n!}{4!(n-4)!}$ are in the ratio $2:1$,then $n=$

If ${ }^{n} C_{12}={ }^{n} C_{8}$,then $n$ is equal to

If $^nC_4$, $^nC_5$ and $^nC_6$ are in arithmetic progression ($A$.$P$.), then the value of $n$ is...

Determine the number of $5$ card combinations out of a deck of $52$ cards if there is exactly one ace in each combination.

If ${}^n C_7 = {}^n C_6$,then ${}^n C_2 = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module