Q.

The number of non-negative solutions of x1+x2+x3+,…,+xn≤n (where n is positive integer) is

see full answer

Talk to JEE/NEET 2025 Toppers - Learn What Actually Works!

Real Strategies. Real People. Real Success Stories - Just 1 call away
An Intiative by Sri Chaitanya

a

2nCn−1

b

2n−1Cn−1

c

2n+1Cn−1

d

2n−1Cn−1−1

answer is A.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

In general, we know thatFor the distribution equationx1+x2+x3+…+xn≤nLet required ways = W⇒W=No. of ways of distributing 1 item+No. of ways of distributing 2 items+......+No. of Ways of distributing n items=1+n−1Cn−1+2+n−1Cn−1+…+n+n−1Cn−1=nCn−1+n+1Cn−1+…+2n−1Cn−1= nCn−1+nCn+n+1Cn−1+…+2n−1Cn−1−nCn= n+1Cn+n+1Cn−1+…+2n−1Cn−1−nCn- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -= 2n−1Cn+2n−1Cn−1−nCn=2nCn−nCn∴ W=2nCn−1
Watch 3-min video & get full concept clarity
score_test_img

Get Expert Academic Guidance – Connect with a Counselor Today!

whats app icon