1 Marks Each

Question 511 Mark

If ${ }^{ n } P _{ r }+{ }^{ n } C _{ r }=24$ then find the value of $r$ .

Answer

$\frac{ n P_r}{ n C_r}=24$
$\therefore \frac{\frac{ n !}{( n - r )!}}{ n !( n - r )!}=24$
$\therefore r! =24$
$\therefore r! =4 \times 3 \times 2 \times 1$
$\therefore r! =4!$
$\therefore r=4$

View full question & answer→

Question 521 Mark

Obtain the value of the following.

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}$
${ }^n C_x={ }^n C_y, x=y \text { OR } n= x + y$
${ }^{27} C_{r+4}={ }^{27} C_{2 r-1}$
$\therefore r+4=2 r-1$
$\therefore 4+1=2 r-r$
$\therefore r=5$
$r+4+2 r-1=27$
$\therefore 3 r+3=27$
$\therefore 3 r=27-3$
$\therefore r=\frac{24}{3}=8$
$(4)$ ${ }^n C_{n-2}=15$
$\therefore \frac{n!}{(n-2)!(n-n+2)!}=15$
$\therefore \frac{n(n-1)(n-2)!}{(n-2)!2!}=15$
$\therefore \frac{n(n-1)}{2}=15$
$\therefore n(n-1)=30$
$\therefore n(n-1)=6 \times 5$
$\therefore n(n-1)=6(6-1)$
$\therefore n=6$

View full question & answer→

Question 531 Mark

If $3 .{ }^{(n+3)} P_4=5 \cdot{ }^{(n+2)} P_4$, then find the value of $n$.

Answer

$3.$ ${ }^{(n+3)} P_4=5 .(n+2) P_4$
$\therefore 3(n+3)(n+3-)(n+3-2)\left[\text { According to definition of }{ }^n P_r\right]$
$=5 \cdot(n+3)(n+2-1)(n+2-2)(n+2-3)$
$\therefore 3 \cdot(n+3)(n+2)(n+1)(n)=5 \cdot(n+2)(n)(n-1)$
$\therefore 3(n+3)=5(n-1)$
$\therefore 3 n+9=5 n-5$
$\therefore 9+5=5 n-3 n$
$\therefore 14=2 n$
$\therefore n=\frac{14}{2}=7$

View full question & answer→

Question 541 Mark

If ${ }^9 P _{ r }=3024$, find the value of $r$ .

Answer

${ }^n P_r=\frac{ n !}{( n - r )!}$
$\therefore{ }^9 P_r=\frac{9!}{(9- r )!}$
$\therefore 3024=\frac{362880}{(9-r)!}$
$\therefore(9- r )!=\frac{362880}{3024}$
$\therefore(9- r )!=120=5!$
$\therefore(9- r )=5$
$\therefore r =9-5=4$
Hence, $r=4$

View full question & answer→

Question 551 Mark

If ${ }^n P _{ 3 }=990$, then find the value of $n$.

Answer

$n P_3=990$
$\therefore n(n-1)(n-2)=990$
$\therefore n(n-1)(n-2)=11 \times 90$
$\therefore n(n-1)(n-2)=11 \times 10 \times 9$
$\therefore n(n-1)(n-2)=11(11-1)(11-2)$
$\therefore n=11$

View full question & answer→

Question 561 Mark

${ }^{10} P _{ 3 }$ Obtain the value of the following.

Answer

${ }^{10} P_3=\frac{10!}{(10-3)!}=\frac{10!}{7!}=\frac{10 \times 9 \times 8 \times 7 \times!}{7!}=10 \times 9 \times 8=720$
$\therefore{ }^{10} P_3=720$
Alternative method:
${ }^{10} P_3=10(10-1)(10-2)=10 \times 9 \times 8=720$
$\therefore{ }^{10} P_3=720$

View full question & answer→

Statistics 1 Marks Each