Obtain the value of the following. - Vidyadip

Question

Obtain the value of the following.

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Answer

$(1)$ ${ }^{11} P_4$
${ }^{ n } C_r=\frac{ n !}{4!(11-4)!}=\frac{11!}{4!7!}=\frac{1 \times 10 \times 9 \times 8 \times 7!}{4 \times 3 \times 2 \times 1 \times 7!}=\frac{7920}{24}=330$
$(2)$ ${ }^{25} C_{23}$
${ }^{ n } C_r=\frac{ n !}{ r !( n - r )!}$
$\therefore{ }^{25} C_{23}=\frac{25!}{23!(25-23)!}=\frac{25!}{23!2!}=\frac{25 \times 24 \times 23!}{23!\times 2 \times 1}=\frac{600}{2}=300$
$(3)$${ }^{27} C_{r+4}={ }^{27} C_{2 r-1}$
${ }^n C_x={ }^n C_y, x=y \text { OR } n= x + y$
${ }^{27} C_{r+4}={ }^{27} C_{2 r-1}$
$\therefore r+4=2 r-1$
$\therefore 4+1=2 r-r$
$\therefore r=5$
$r+4+2 r-1=27$
$\therefore 3 r+3=27$
$\therefore 3 r=27-3$
$\therefore r=\frac{24}{3}=8$
$(4)$ ${ }^n C_{n-2}=15$
$\therefore \frac{n!}{(n-2)!(n-n+2)!}=15$
$\therefore \frac{n(n-1)(n-2)!}{(n-2)!2!}=15$
$\therefore \frac{n(n-1)}{2}=15$
$\therefore n(n-1)=30$
$\therefore n(n-1)=6 \times 5$
$\therefore n(n-1)=6(6-1)$
$\therefore n=6$

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