Question
Question: Find the value of ‘r’ for which. \[{20_{{{\text{C}}_{\text{r}}}}}{20_{{{\text{C}}_{\text{o}}}}} + ...
Find the value of ‘r’ for which.
20Cr20Co+20Cr−120C1+20Cr−220C2+…+20Co20Cr is maximum.
A) 20
B) 10
C) 15
D) 11
Solution
Here, we will be using the Vandermonde’s formula to solve the question’s first part i.e. first we will find the value of the expression and then we will find the value of ‘r’ for which the expression attains a maximum.
Complete step by step solution: Formula used: Vandermonde’s Formula:
This is the direct formula which we can apply to solve the problem. It says:-
nCrnCo+nCr−1nC1+nCr−2nC2+…+nConCr=2nCr
Here one most remember following points:-
a) nCrnCo+nCr−1nC1+nCr−2nC2+…+nConCr=2nCr n is same in all the terms
b) base terms should add upto same number
here r + o = r | r − 1 + 1 = r | r − 2 + 2 = r | … o + r |
c) and the sum of expression is 2×nCr i.e
2 × power Csum of bases.
So, if we substitute the expression in the Vandermonde’s formula, we get
20Cr20Co+20Cr−120C1+20Cr−220C2+…+20Co20Cr
= 2×20Cr
= 40Cr
Now 40Cr attains maximum at r=240=20.
(Note:- nCr attains maximum at r=2n if n is even
and r=2n+1 if n is odd)
∴ Correct answer is (A) 20.
Note: One must be very careful in applying the Vandermonde’s formula as one should also check that it contains all the terms i.e from nCo nCr … nCr nCo.
If any one of the term is even missing, you cannot apply the formula.