Combination

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Combination

by sumasajja » Wed Aug 03, 2011 8:06 am
Combination
7 identical balls have to be distributed to 4 people such that each one gets at least 1 ball.What are the number of ways of the distribution such that atleast 1 gets 3 balls?

A) 7 B) 28 c) 81 d)2401 e) none of these
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by sumasajja » Wed Aug 03, 2011 12:24 pm
the answer to the question is E but i cant understand the procedure ,,,how it can be done.....please any1 explain the procedure in detail.thank you

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by tpr-becky » Wed Aug 03, 2011 12:31 pm
if you have 4 people and distribute 7 balls so that each person gets at least 1 ball and at least 1 person gets 3 balls you only have 1 extra ball to hand out after each person gets one.

That extra ball could go to the person with three - in which case you would have 4111 (but there are 4 places for the 4 to go: 4111, 1411, 1141, 1114)

If the extra ball goes to a person with 1 ball then you have 3,2,1,1 and you have to figure out how many ways you can order this. one easy wat is to look for a pattern and follow it methodically:

Starts with 3: 3211, 3121, 3112
Starts with 2: 2311, 2131, 2113
starts with 1: 1123, 1132, 1213, 1231, 1312, 1321

for 16 total ways.
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by gmatboost » Wed Aug 03, 2011 9:36 pm
This question is written very poorly:

1. The writing contains errors
2. Does "At least 1 person gets 3 balls" mean "at least 3" or "exactly 3"?
3. The GMAT would be very unlikely to say "At least 1 person gets 3 balls" when only 1 person can possibly get that many

I would be careful about using questions from the source of this question.
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by Prashant Ranjan » Wed Aug 03, 2011 11:11 pm
I agree with Greg that the question is inherently flawed when it says that at least 1 person should get 3 balls when actually only 1 person can get 3 balls or more.

Consider the four persons as P1,P2,P3,P4. Each gets 1 ball which can be distributed as 4C4 ways.

P1----P2----P3----P4
1-----1-----1-----1

Now 1 person can get 3 balls when the remaining 2 balls are distributed to any of the 4 persons as
4C1.

P1----P2----P3----P4
1-----1-----1-----1
1-----1
1

The remaining 1 ball can be distributed to remaining 3 persons in 3C1 ways as shown above.

Hence the net chances are 4C4 * 4C1 * 3C1 = 12 ways

However there can be one more case of distributing the 3 remaining balls so that all three balls are assigned to 1 person among the 4 persons.

P1----P2----P3----P4
1-----1------1-----1
1
1
1

This can be done on 4C4 * 4C1 ways = 4
Hence net chances are 12 + 4 = 16 ways

However while solving this, I have assumed that the question asks how can the 7 balls be distributed so that 1 person gets at least 3 balls.

Thanks

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by GMATGuruNY » Thu Aug 04, 2011 2:52 am
sumasajja wrote:Combination
7 identical balls have to be distributed among 4 people. How many ways can the balls be distributed so that each person gets at least 1 ball and one person gets more than 2 balls?

A) 7 B) 28 c) 81 d)2401 e) none of these
The approaches above are great. Below is one more.

One person gets 3 balls:
In order for each person to get at least 1 ball, the balls must be distributed 3-2-1-1.
Number of ways to arrange the 4 elements {3,2,1,1} = 4!/2! = 12.

One person gets 4 balls:
In order for each person to get at least 1 ball, the balls must be distributed 4-1-1-1.
Number of ways to arrange the 4 elements {4,1,1,1} = 4!/3! = 4.

Total possible distributions = 12+4 = 16.

The correct answer is E.
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