{ }^{15}P_8 = A + 8 cdot { }^{14}P_7 Rightarr | vedclass

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

Question

(B) Given,${ }^{15}P_8 = A + 8 \cdot { }^{14}P_7$ We know that ${ }^{n}P_r = \frac{n!}{(n-r)!}$. So,$\frac{15!}{(15-8)!} = A + 8 \cdot \frac{14!}{(14-

Vedclass · Accepted Answer

${ }^{14}P_8$

Vedclass · Answer

(B) Given,${ }^{15}P_8 = A + 8 \cdot { }^{14}P_7$ We know that ${ }^{n}P_r = \frac{n!}{(n-r)!}$. So,$\frac{15!}{(15-8)!} = A + 8 \cdot \frac{14!}{(14-7)!}$ $\frac{15!}{7!} = A + 8 \cdot \frac{14!}{7!}$ $A = \frac{15!}{7!} - 8 \cdot \frac{14!}{7!}$ $A = \frac{15 \cdot 14!}{7!} - 8 \cdot \frac{14!}{7!}$ $A = \frac{14!}{7!} (15 - 8)$ $A = \frac{14!}{7!} (7)$ $A = \frac{14!}{6!} = { }^{14}P_8$

${ }^{15}P_8 = A + 8 \cdot { }^{14}P_7 \Rightarrow A = $

Similar Questions

Which of the following is correct for $Chlorophyceae$?

Which of the following reagents will convert propionic acid to $1-propanol$?

In which of the following reactions is the product an ether?

$\sec h^{-1}\left(\frac{1}{2}\right)-\operatorname{cosec} h^{-1}\left(\frac{3}{4}\right)$ is equal to

Let $A = \{ x: x \in R, |x| < 1 \}$ and $B = \{ x: x \in R, |x - 1| \ge 1 \}$. If $A \cup B = R - D$,then the set $D$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module