JEE Main 2023 · 11 April · morning · Single correct · +4 / −1
The number of triplets (x, y, z), where x, y, z are distinct non negative integers satisfying x+y+z=15, is :
- (a) 136
- (b) 80
- (c) 92
- (d) 114
Answer and solution
Answer (d) 114
Solution
Maths sir · 4 steps
Count all solutions first, then throw out the ones where two or more of x, y, z are equal.
- All solutions. With no distinctness condition, the non-negative solutions of x+y+z=15 number C(15+2, 2) = C(17, 2) = 136.
- All three equal. x=y=z needs 3x=15, so only (5,5,5) works. That is 1 solution.
- Exactly two equal. Take x=y, so 2x+z=15 with x = 0, 1, …, 7. That gives 8 solutions, but x=5 gives (5,5,5), so 7 have exactly two equal. The equal pair can be (x,y), (y,z) or (z,x), giving 3 × 7 = 21.
136 - 21 - 1 = 114
- So. There are 114 ordered triplets with x, y, z all distinct, which is option (d).
The idea to keep. For distinct values, use stars and bars for the total, then subtract the cases with exactly two equal and the case with all three equal. Count each case only once.
Answer from the NTA final answer key. Solution by PrepWiser.