# Combinatorics Doubt

while calculating nCr (repetition is allowed) , I found that if r>n/2 then we can simply write r = n-r. Can someone please explain why or provide some links where i can find it. Thanks!

1 Like

nCr = nCn-r . Selecting r objects out of n is same as selecting n-r objects and throwing them away. You end up having r objects.

1 Like

ncr is number of ways of choosing r objects out of n
by choosing r elements…u have automatically chosen n-r elements
by choosing who is in ur team …u have automatically chosen who is not in ur team
so C(n,r)=C(n,n-r)

so if r>n/2 then n-r is smaller than r

so people calculate that

this is a bad proof but u can prove mathematically

3 Likes

https://youtu.be/KbB0FjPg0mw watch last 15 minutes

1 Like

Write the expression for ^nC_{n-r} in expanded form and you will get ^nC_{r} as shown below:

^nC_{n-r} = \frac{n!} {(n-r)! \times (n - (n-r))!} = \frac{n!} {(n - r)! \times r!} = ^nC_{r}

1 Like

First of all, for nCr, it doesn’t matters that elements are distinct or repeated.
This is only The number of ways to choose “r” elements out of “n” elements.
for example consider an sequence X = {a,b,c,d}, here n = 4 and u are asked that how many ways u can select 2 elements from this array, then your answer will be = 4C2 = 6.
these are the possible ways : {1,2}, {1,3}, {1,4}, {2,3}, {2,4}, {3,4}
Note that above pairs do not describe about properties of elements, instead these are indexes (say first and second, first and third…like this).
The mathematical expression of nCr = n! / (r! *(n-r)!) and nC(n-r) = n! / (r!* (n-r)!).
Both are same.

1 Like