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In how many ways can we arrange 7 different things to 3 people. All of them must get at least one.

My Approach

I used the formula (n-$1$)C(r-$1$)=$6$C$2$=$15$

But I am confused will it work for different things also?

N. F. Taussig
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justin takro
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    You are right, you cannot use the formula for different things. – cr001 Nov 16 '15 at 06:26
  • This does not work. I would be tempted to use Inclusion/Exclusion. There is a more awkward approach using cases, and a fancier approach using Stirling numbers. – André Nicolas Nov 16 '15 at 06:28
  • @AndréNicolas Can you guide how to solve.As i was unable to guide by your hint.What is exactly inclusive/exclusive principle.Where exactly I can apply these? – justin takro Nov 17 '15 at 14:40
  • From $3^7$, subtract the $\binom{3}{1}2^7$ ways to give to two people, but this overcompensates for the ways to give all to one. Total $3^7-\binom{3}{1}2^7+\binom{3}{2}$. – André Nicolas Nov 17 '15 at 15:11
  • @AndréNicolas I have not understood your question? – justin takro Nov 17 '15 at 16:21
  • There is an accepted answer already, which is fully correct in outline, with some of the numerical work omitted. The number of "bad" ways to distribute (one or two people get nothing) is $\binom{3}{1}2^7-\binom{3}{2}1^7$. So the final answer is $2187-384+3$. – André Nicolas Nov 17 '15 at 16:34

1 Answers1

3

Hint:

Each object has 3 options, as each object can go to either of the three objects. This means that there are a total of $3^7$ options. Now, subtract the cases in which there is at least one person with zero objects.

Gummy bears
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  • I think that would be 1 case with zero objects.So,Ans would be $3^7$ -$1$. – justin takro Nov 16 '15 at 06:36
  • @justintakro Think again! There are 3 people that can have zero objects, one at a time. Moreover, they can also have zero objects 2 at a time. Meaning that there can be 3 cases when 1 person has zero objects. How many cases are there when 2 people have zero objects? – Gummy bears Nov 16 '15 at 07:21
  • "There are 3 people that can have zero objects, one at a time"Does this mean either A can have(then B and C does not have) or B can have(then A and C does not have) or C can have(then B and A does not have).Either A and B do not have (C has),B and C do not have(A has),A and C do not have(B has). – justin takro Nov 17 '15 at 14:36
  • @Grummy bears Ans given is $381$.I do not get correct Ans here. – justin takro Nov 17 '15 at 14:39
  • @justintakro That answer seems incorrect... Are you sure it's not 2181? – Gummy bears Nov 17 '15 at 15:29