For 2 le r le n,the expression binom{n}{r} + | vedclass

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

Question

(D) The given expression is $\binom{n}{r} + 2\binom{n}{r-1} + \binom{n}{r-2}$.We can rewrite the middle term $2\binom{n}{r-1}$ as $\binom{n}{r-1} + \b

Vedclass · Accepted Answer

$\binom{n+2}{r}$

Vedclass · Answer

(D) The given expression is $\binom{n}{r} + 2\binom{n}{r-1} + \binom{n}{r-2}$.We can rewrite the middle term $2\binom{n}{r-1}$ as $\binom{n}{r-1} + \binom{n}{r-1}$.So,the expression becomes $\binom{n}{r} + \binom{n}{r-1} + \binom{n}{r-1} + \binom{n}{r-2}$.Using the Pascal's identity $\binom{n}{k} + \binom{n}{k-1} = \binom{n+1}{k}$,we get:$(\binom{n}{r} + \binom{n}{r-1}) + (\binom{n}{r-1} + \binom{n}{r-2}) = \binom{n+1}{r} + \binom{n+1}{r-1}$.Applying the identity again,$\binom{n+1}{r} + \binom{n+1}{r-1} = \binom{n+2}{r}$.Thus,the correct option is $(d)$.

For $2 \le r \le n$,the expression $\binom{n}{r} + 2\binom{n}{r-1} + \binom{n}{r-2}$ is equal to:

Similar Questions

$\binom{n}{n-r} + \binom{n}{r+1}$,whenever $0 \le r \le n-1$,is equal to

There are $5$ gentlemen and $4$ ladies to dine at a round table. In how many ways can they seat themselves so that no $2$ ladies are together?

How many integers between $1000$ and $10000$ have no digits other than $4, 5,$ or $6$?

The number of ways in which the letters of the word $ARRANGE$ can be arranged such that both $R$ do not come together is

How many numbers lying between $1000$ and $10000$ can be formed by using the digits $1, 3, 5, 6, 7, 8, 9$ if no digits are repeated?

Vedclass Test Series

Exam Paper Generator

Online Exam Module