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Number of ways in which 9 different toys can be distributed among 4 ch
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Bunuel
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Number of ways in which 9 different toys can be distributed among 4 ch [#permalink] New post  10 Feb 2021, 08:00
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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) \(\frac{9!}{3!(2!)^4}\)


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Gknight5603
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Re: Number of ways in which 9 different toys can be distributed among 4 ch [#permalink] New post  10 Feb 2021, 08:40
AnsD. 9c3*6c2*4c2*2c2

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wishmasterdj
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Re: Number of ways in which 9 different toys can be distributed among 4 ch [#permalink] New post  12 Feb 2021, 05:22
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
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