If sumlimits_{r = 0}^{25} {left( {^{50}C_r cd | vedclass

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

Question

(D) We know that $^{n}C_{r} \cdot ^{n-r}C_{k-r} = \frac{n!}{r!(n-r)!} \cdot \frac{(n-r)!}{(k-r)!(n-k)!} = \frac{n!}{r!(k-r)!(n-k)!}$.Multiplying and d

Vedclass · Accepted Answer

$2^{25}$

Vedclass · Answer

(D) We know that $^{n}C_{r} \cdot ^{n-r}C_{k-r} = \frac{n!}{r!(n-r)!} \cdot \frac{(n-r)!}{(k-r)!(n-k)!} = \frac{n!}{r!(k-r)!(n-k)!}$.Multiplying and dividing by $k!$,we get $\frac{n!}{k!(n-k)!} \cdot \frac{k!}{r!(k-r)!} = ^{n}C_{k} \cdot ^{k}C_{r}$.Applying this to the given sum:$\sum\limits_{r = 0}^{25} {^{50}C_r \cdot ^{50 - r}C_{25 - r}} = \sum\limits_{r = 0}^{25} {^{50}C_{25} \cdot ^{25}C_r}$.$= ^{50}C_{25} \sum\limits_{r = 0}^{25} {^{25}C_r}$.Since $\sum\limits_{r = 0}^{n} {^{n}C_r} = 2^n$,we have $\sum\limits_{r = 0}^{25} {^{25}C_r} = 2^{25}$.Therefore,the expression becomes $^{50}C_{25} \cdot 2^{25}$.Comparing this with $K\left( {^{50}C_{25}} \right)$,we get $K = 2^{25}$.

If $\sum\limits_{r = 0}^{25} {\left( {^{50}C_r \cdot ^{50 - r}C_{25 - r}} \right) = K\left( {^{50}C_{25}} \right)}$,then $K$ is equal to

Similar Questions

The number of $3$-digit numbers,formed using the digits $2, 3, 4, 5$ and $7$,when the repetition of digits is not allowed,and which are not divisible by $3$,is equal to ..........

Numbers are to be formed between $1000$ and $3000$,which are divisible by $4$,using the digits $1, 2, 3, 4, 5$ and $6$ without repetition of digits. Then the total number of such numbers is.

If all the letters of the word $ACADEMICIAN$ are permuted in all possible ways,then the number of permutations in which no two $A$'s are together and all the consonants are together is:

$9$ balls are to be placed in $9$ boxes. $3$ boxes are so small that they cannot hold $5$ balls. In how many ways can one ball be placed in each box?

The number of even numbers greater than $1000000$ that can be formed using all the digits $1, 2, 0, 2, 4, 2, 4$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module