#Discrete #Combinatorics

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The number of ways in which n distinct objects can be put into two identical boxes so that no box remains empty, is

a) 2^n - 1

b) 2^n - 2

c) 2^(n-1) - 1

d) None of these

retagged
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is it c? i just try with example
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yes the ans is c.

S(n,2) =  2^(n-1) - 1

http://www.careerbless.com/aptitude/qa/permutations_combinations_imp7.php

edited by
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Got it.2^n - 2 ways if the boxes are not identical. But why did you subtract one for identical boxes case.Shouldn't we divide by two for identical boxes case to counter case which now will be similar due to identical boxes.
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I have edited the answer. See.
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Doubt in a similar question.

How many ways are there to distribute five distinguishable objects into three indistinguishable boxes?

Can we solve it like following?

First, we calculate the no of ways of distributing 5 distinct objects in 3 distinct boxes which are = $3^{5}$ and then we divide it with 3! as we can label the boxes in 3! ways.

Final answer=$3^{5}$ /3!

I know that using the formula of Stirling numbers of the second kind, it can be solved as S(5,1)+S(5,2)+S(5,3). But can the method stated above be used?

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In S(k,n) what is S.How do we calculate such expression?
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In S(k,n), S is the Stirling number of the second kind

Stirling numbers of the second kind obey the recurrence relation

S(k,n)= 1, if k>0 and n=1

1, if k=n

n*S(k-1,n)+S(k-1,n-1), 0<=n<=k

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