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CAT 2017 Slot 1 — QA Question 32

Letters and Letter BoxesEasy

In how many ways can 7 identical erasers be distributed among 4 kids in such a way that each kid gets at least one eraser but nobody gets more than 3 erasers?

Answer & solution

Correct answer: 16

  • 16

  • B

    20

  • C

    14

  • D

    15

Solution

Easy

Give each kid one eraser first, leaving 33 to distribute. The cap "no more than 33" means no kid may receive 44 in total, i.e. no one may take all 33 extras. Count the unrestricted distributions of the 33 extras, then remove the cases that break the cap.

1

Reduce to distributing the leftovers. After 11 each, 7−4=37-4=3 erasers remain among 44 kids (w′,x′,y′,z′≥0w',x',y',z'\ge 0).

w′+x′+y′+z′=3\begin{aligned} &w' + x' + y' + z' = 3 \end{aligned}
2

Count all non-negative solutions (stars and bars).

(3+4−14−1)=(63)=20\begin{aligned} &\binom{3+4-1}{4-1} = \binom{6}{3} = 20 \end{aligned}
3

Remove cap violations. A kid exceeds 33 total only by taking all 33 extras (becoming 44). That happens in exactly 44 ways, one per kid.

20−4=16\begin{aligned} &20 - 4 = 16 \end{aligned}

Directly split the 33 extras: either 2+12{+}1 to two kids — 4P2=12^4P_2 = 12 ways — or 1+1+11{+}1{+}1 to three kids — (43)=4\binom{4}{3}=4 ways. Total 12+4=1612+4=16.

16 ways16 \text{ ways}

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CAT 2017 Slot 1 QA Q32: In how many ways can 7 identical erasers be distributed among 4 kids in such a way that each kid gets at least — Solution | TheCATExam