^nC_r + ^nC_{r-1} is equal to | vedclass

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

Question

(A) The expression $^nC_r + ^nC_{r-1}$ is a standard identity in combinatorics known as Pascal's Identity.Using the formula $^nC_r = \frac{n!}{r!(n-r)

Vedclass · Accepted Answer

$^{n+1}C_r$

Vedclass · Answer

(A) The expression $^nC_r + ^nC_{r-1}$ is a standard identity in combinatorics known as Pascal's Identity.Using the formula $^nC_r = \frac{n!}{r!(n-r)!}$,we have:$^nC_r + ^nC_{r-1} = \frac{n!}{r!(n-r)!} + \frac{n!}{(r-1)!(n-r+1)!}$$= \frac{n!}{r(r-1)!(n-r)!} + \frac{n!}{(r-1)!(n-r+1)(n-r)!}$$= \frac{n!}{(r-1)!(n-r)!} \left( \frac{1}{r} + \frac{1}{n-r+1} \right)$$= \frac{n!}{(r-1)!(n-r)!} \left( \frac{n-r+1+r}{r(n-r+1)} \right)$$= \frac{n!}{(r-1)!(n-r)!} \left( \frac{n+1}{r(n-r+1)} \right)$$= \frac{(n+1)n!}{r(r-1)! (n-r+1)(n-r)!}$$= \frac{(n+1)!}{r!(n+1-r)!} = ^{n+1}C_r$Thus,the correct option is $A$.

$^nC_r + ^nC_{r-1}$ is equal to

Similar Questions

There are $4$ oranges,$5$ apples,and $7$ mangoes in a fruit basket. The number of ways of selecting at least one fruit from among the fruits in the basket is

If ${}^nC_{r-1}=330$,${}^nC_r=462$,and ${}^nC_{r+1}=462$,then $r$ is equal to

How many distinct combinations can be formed using one or more letters from the word $MISSISSIPPI$?

In a lottery,a person chooses six different natural numbers at random from $1$ to $20$. If these six numbers match the six numbers already fixed by the lottery committee,he wins the prize. What is the probability of winning the prize in the game? [Hint: The order of the numbers is not important.]

Let the number of elements in sets $A$ and $B$ be $5$ and $2$ respectively. Then the number of subsets of $A \times B$ each having at least $3$ and at most $6$ elements is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module