Question bank/JEE Main/Mathematics

JEE MainMathematicsModeratePermutations and Combinations

The number of ways in which 15 identical toys can be distributed
among 3 children so that each of them gets at least two toys is equal to

Show answer and solution
x1+x2+x3=15x_{1} + x_{2} + x_{3} = 15
xi≥2, let yi=xi−2x_{i} \geq 2,\ let\ y_{i} = x_{i} - 2
⇒y1+y2+y3=9,yi≥0\Rightarrow y_{1} + y_{2} + y_{3} = 9,y_{i} \geq 0
∴\therefore no. of ways = 11C2= \ ^{11}C_{2}

Source: Grass-G / M-08 Mathematics Paper 23 October.docx

More JEE Main questionsChapter practice

Need a video for this idea?

Scan any doubt from your notes and get a whiteboard explanation.

Scan a doubtView plans