PC

There are three sets of objects.
One group contains n elements of one kind

Other contains n elements of another kind
Last grp contains n different elements

All 3n elements are mixed together.

Find the number of ways in which n things can be selected from this pile

4 Answers

\sum_{k=0}^{n}{\left( \sum_{r=0}^{n}{\left \binom{n}{(k+r)} }\right)} ?

106

Asish Mahapatra ·2010-02-06 02:42:42

ans: (n+2).2^n-1

yup the answer is same as the one given by asish... :)

If we chose r objects from the first 2 groups, the number of ways to do that is r+1

Now we have to chose n-r from the last group which can be done in ⁿC_n-r ways

so the total number of ways is

\sum_{r=0}^{n}{(r+1)\times^nC_r}=\sum_{r=0}^{n}{(r)\times^nC_r+^nC_r}=n.2^{n-1}+2^n

106

Asish Mahapatra ·2010-02-07 21:30:56

thanks sir and akari..

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