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1)7 different objects must be divided among 3 people.In how

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1)7 different objects must be divided among 3 people.In how [#permalink]

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New post 24 Jun 2004, 11:07
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(1)7 different objects must be divided among 3 people.In how many ways can this be done if one or tow of them can get no objects?
(a)15120 (b)2187 (c)3003 (d)792



(2)7 different objects must be divided among 3 people.In how many ways can this be done if at least one out of them must get no objects?
(a)381 (b)36 (c)84 (d)174



(3)7 different objects must be divided among 3 people.In how many ways can this be done if at least one of them gets exactly one object?
(a)2484 (b)1176 (c)729 (d)None of these
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New post 25 Jun 2004, 00:12
hello wannabe

welcome to gmatclub. you will have a better response if you post one question per thread.
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Re: Permutation & Combination [#permalink]

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New post 25 Jun 2004, 00:50
wanna_mba_2005 wrote:
(1)7 different objects must be divided among 3 people.In how many ways can this be done if one or tow of them can get no objects?
(a)15120 (b)2187 (c)3003 (d)792



(2)7 different objects must be divided among 3 people.In how many ways can this be done if at least one out of them must get no objects?
(a)381 (b)36 (c)84 (d)174



(3)7 different objects must be divided among 3 people.In how many ways can this be done if at least one of them gets exactly one object?
(a)2484 (b)1176 (c)729 (d)None of these


1. In 3^7 = 2187 ways, because each object can go to 3 different people, and by the rule of multiplying we get 3*3*...*3 = 3^7 ways.

2. If first has no objects, then 2^7. => 3*2^7 - (common cases, when at least 2 get no objects, there are 3 such cases) = 128*3 - 3 = 381.
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Re: Permutation & Combination [#permalink]

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New post 25 Jun 2004, 00:58
wanna_mba_2005 wrote:
(3)7 different objects must be divided among 3 people.In how many ways can this be done if at least one of them gets exactly one object?
(a)2484 (b)1176 (c)729 (d)None of these


# of combs where all three get exactly 1 object is 0.

# of combs where 2 get exactly 1 object is 3*2*C[2,7].

# of combs where 1 gets exactly 1 object is 3*7*(2^6 - 6).

=> total # = 6*7*6/2 + 3*7*58 = 7*(18 + 174) = 192*7 = 1344.

None of the answers... Maybe it is reasonable to notice that 7 should divide the answer. 2484 does not satisfy, 729 either. But 1176 = 24*(50-1) :: 7...
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Permutation & Combination [#permalink]

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New post 25 Jun 2004, 11:05
Thanks to all of you for solving the problems.The third one still continues to nag me.The official answer happens to be (b) and I am clueless...just one of those things you can sleep over!
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Re: 1)7 different objects must be divided among 3 people.In how [#permalink]

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New post 15 Nov 2014, 14:00
The answer is 2187.

"One or two of them can get no objects" means objects can be distributed in any way, that is the first person can either get an object or not, the 2nd person can either get an object or not, and the 3rd person can either get an object or not. The task is the same as if we were simply asked t calculate the number of ways in which 3 objects can be distributed among 7 people and each person should not necessarily get an object.

Thus, the number of ways is 3^7=2187
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Re: 1)7 different objects must be divided among 3 people.In how [#permalink]

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Re: 1)7 different objects must be divided among 3 people.In how   [#permalink] 12 Feb 2016, 16:06
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