# CAT 2021 Set-2 | Quantitative Aptitude | Question: 4

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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

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First, we can satisfy the minimum requirement of three children names $C_{1},C_{2}, \text{and} \; C_{3}.$
$$\begin{array}{|c|c|c|} \hline C_{1} & C_{2} & C_{3} \\\hline \text{4 Balloons} & \text{4 Balloons} & \text{4 Balloons} \\ + & + & + \\ \text{1 Pencil} & \text{1 Pencil} & \text{1 Pencil} \\\hline \end{array}$$
Now, we are left with $3 \; \text{Balloons}, 3 \; \text{Pencils}, \text{and} \;3 \; \text{Erasers}.$

$3 \; \text{Balloons}$ we can distribute among three children, in that way.

$C_{1}+C_{2}+ C_{3}=3\;; C_{1},C_{2},C_{3}>0$

The number of ways $=\;^{3+3-1}C_{3-1} = \;^{5}C_{2} = 10$

Similarly, we can distribute $3 \; \text{Pencils}, \text{and} \; 3 \; \text{Erasers}.$

$\therefore$ The total number of ways $= 10 \times 10 \times 10 = 1000$ ways.

Correct Answer $:1000$

$\textbf{PS:}$

• The number of non-negative integral solutions of equations $x_{1} + x_{2} + x_{3} + \dots + x_{r} = n\;,\;x_{i}>0,i=1,2,3,\dots, r$ is $\binom{n+r-1}{n} = \binom{n+r-1}{r-1}$ ways.
• In general, the same analysis shows that the number of ways to distribute $n$ candies (identical) among $r$ kids (distinct) is $\binom{n+r-1}{n},$ or equivalently $\binom{n+r-1}{r-1}.$
• The number of ways to place $n$ indistinguishable (identical) balls into $r$ labelled (distinct, distinguishable) urns is $\binom{n+r-1}{n} = \binom{n+r-1}{r-1}.$
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