The number of ways of choosing 10 objects out of 31 objects of which 10 are identical and the remaining 21 are distinct, is
- A.
- B.
- C.
- D.
Show answer and solution
Correct option: C
Solution
If we choose r identical objects (r = 0,1,...,10), we must choose (10−r) from the 21 distinct ones, giving C(21, 10−r) ways. Total = Σ(r=0 to 10) C(21, 10−r) = C(21,10) + C(21,9) + ... + C(21,0).
By symmetry of binomial coefficients, C(21,k) = C(21,21−k), so this sum equals C(21,11) + C(21,12) + ... + C(21,21), which is exactly half of the full sum 2²¹ (since the sum of all C(21,k) = 2²¹, and the middle term C(21,10.5) doesn't exist — 21 is odd so there's no middle term, meaning the sum splits evenly into two equal halves each = 2²⁰).
Therefore the answer is 2²⁰.
By symmetry of binomial coefficients, C(21,k) = C(21,21−k), so this sum equals C(21,11) + C(21,12) + ... + C(21,21), which is exactly half of the full sum 2²¹ (since the sum of all C(21,k) = 2²¹, and the middle term C(21,10.5) doesn't exist — 21 is odd so there's no middle term, meaning the sum splits evenly into two equal halves each = 2²⁰).
Therefore the answer is 2²⁰.
[JEE Main 2019 · 12 Apr · Shift I]