TheCATExam
Sign in

CAT 2021 Slot 2 — QA Question 6

Letters and Letter BoxesEasy

The number of ways of distributing 15 identical balloons, 6 identical pencils and 3 identical erasers among 3 children, such that each child gets at least four balloons and one pencil, is 

Answer & solution

Answer: 1000

Solution

Easy

Each item is independent, so distribute them separately and multiply the counts. First hand out the compulsory minimums, then distribute the leftover identical items freely using stars-and-bars.

Distributing nn identical items among rr children (each child may get 00) has (n+r−1r−1)\dbinom{n+r-1}{r-1} ways. Here r=3r=3.

1

Balloons. 1515 balloons, each child must get at least 44. Give 44 to each (1212 used), leaving 33 to distribute freely.

(3+3−13−1)=(52)=10 ways\begin{aligned} &\binom{3+3-1}{3-1}=\binom{5}{2}=10\ \text{ways} \end{aligned}
2

Pencils. 66 pencils, each child must get at least 11. Give 11 to each (33 used), leaving 33 free.

(3+3−13−1)=(52)=10 ways\begin{aligned} &\binom{3+3-1}{3-1}=\binom{5}{2}=10\ \text{ways} \end{aligned}
3

Erasers. 33 erasers with no minimum, distributed freely.

(3+3−13−1)=(52)=10 ways\begin{aligned} &\binom{3+3-1}{3-1}=\binom{5}{2}=10\ \text{ways} \end{aligned}
4

Combine. The three distributions are independent, so multiply.

10×10×10=1000\begin{aligned} &10\times 10\times 10=1000 \end{aligned}
Number of ways=1000\text{Number of ways}=1000
CAT 2021 Slot 2 QA Q6: The number of ways of distributing 15 identical balloons, 6 identical pencils and 3 identical erasers among 3 — Solution | TheCATExam