Question

If ${C_r} = {}^{25}{C_r}$and ${C_0} + 5.{C_1} + 9.{C_2} + ..... + 101.{C_{25}} = {2^{25}}.k$ then $k$ is equal to

Answer 1

Hint: In this question, we are given with the sum ${C_0} + 5.{C_1} + 9.{C_2} + ..... + 101.{C_{25}} = {2^{25}}.k$ where the value of ${C_r} = {}^{25}{C_r}$ (given). First step is to put the value of ${C_r}$ in the equation of given sum. Then reverse the term of the sum. In last add the given equation of the sum and the equation which we required after reversing all the term. In last you’ll get the sum in terms of ${2^{25}}.k$ so just compare it with the final answer.

Formula Used:
Property of combination formula –
${}^n{C_r} = {}^n{C_{n - r}}$

Complete step by step Solution:
Given that,
$S = {C_0} + 5.{C_1} + 9.{C_2} + .....97{C_{24}} + 101.{C_{25}} = {2^{25}}.k - - - - - \left( 1 \right)$
And ${C_r} = {}^{25}{C_r} - - - - - \left( 2 \right)$
From equation (1) and (2)
$S = {}^{25}{C_0} + 5.{}^{25}{C_1} + 9.{}^{25}{C_2} + .....97{}^{25}{C_{24}} + 101.{}^{25}{C_{25}} - - - - - \left( 3 \right)$
Reverse all the terms of equation (3)
$S = 101.{}^{25}{C_{25}} + 97{}^{25}{C_{24}} + ..... + 9.{}^{25}{C_2} + 5.{}^{25}{C_1} + {}^{25}{C_0} = {2^{25}}.k$
Using the property ${}^n{C_r} = {}^n{C_{n - r}}$ in all the co-efficients
$S = 101.{}^{25}{C_0} + 97{}^{25}{C_1} + ..... + 9.{}^{25}{C_{23}} + 5.{}^{25}{C_{24}} + {}^{25}{C_{25}} - - - - - \left( 4 \right)$
Adding equation (3) and (4)
$2S = 102\left( {{}^{25}{C_0} + {}^{25}{C_1} + ..... + {}^{25}{C_{23}} + {}^{25}{C_{24}} + {}^{25}{C_{25}}} \right)$
$2S = 102\left( {{2^{25}}} \right)$
From equation (1),
$S = 51\left( {{2^{25}}} \right) = k{.2^{25}}$
$ \Rightarrow k = 51$
Hence, the value of $k$ is $51$.

Note: The key concept involved in solving this problem is the good knowledge of combination and permutation. Students must remember that a combination is a collection of $r$ items chosen without repetition from a collection of $n$ items in any order. The key distinction between a combination and a permutation is that order is irrelevant. Order is important in permutations. Also, to solve such question students know that the sum of all combination of $n$different thing is ${2^n}$or we can say that ${}^n{C_0} + {}^n{C_1} + {}^n{C_2} + {}^n{C_3} + ...... + {}^n{C_n} = {2^n}$.