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10 Feb 2021, 09:00
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D
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Difficulty:
45%
(medium)
Question Stats:
72% (02:35) correct
28%
(02:56)
wrong
based on 92
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Number of ways in which 9 different toys can be distributed among 4 children belonging to different age groups in such a way that distribution among the 3 elder children is even and the youngest one is to receive one toy more, is:
(A) \(\frac{(5!)^2}{8}\)
(B) \(\frac{9!}{2}\)
(C) \(\frac{9!}{3!(2!)^2}\)
(D) \(\frac{9!}{3!(2!)^3}\)
(E) \(\frac{9!}{3!(2!)^4}\)
Are You Up For the Challenge: 700 Level Questions: 700 Level Questions
(A) \(\frac{(5!)^2}{8}\)
(B) \(\frac{9!}{2}\)
(C) \(\frac{9!}{3!(2!)^2}\)
(D) \(\frac{9!}{3!(2!)^3}\)
(E) \(\frac{9!}{3!(2!)^4}\)
Are You Up For the Challenge: 700 Level Questions: 700 Level Questions
10 Feb 2021, 09:40
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AnsD. 9c3*6c2*4c2*2c2
Posted from my mobile device
Posted from my mobile device
wishmasterdj
Joined: 04 May 2016
Last visit: 25 Oct 2021
Posts: 91
Own Kudos:
Given Kudos: 10
Location: India
Schools: ISB '18 (A)
GMAT 1: 700 Q48 V37
GPA: 3.2
12 Feb 2021, 06:22
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order not imp as the AB packet is the same as BA (even though the toys are different, but after it becomes a packet of 2 or 3, the packet is the unit in which the order doesn't matter to you)
so, 9C2 * 7C2 * 5C2 * 3C3
so, 9C2 * 7C2 * 5C2 * 3C3
