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Yet another Combinatorics

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Yet another Combinatorics

by Deepthi Subbu » Sat Nov 27, 2010 1:55 am
Gordon buys 5 dolls for his 5 niece . The gifts include 2 identical Sun and Fun beach dolls , one Elegant eddie dress doll , one GI jose doll , and one tulip troll doll . If th eyoungest niece doesnot want the GI jose doll , in how many different ways can he give the gifts?

OA 48
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by Rahul@gurome » Sat Nov 27, 2010 4:22 am
Deepthi Subbu wrote:Gordon buys 5 dolls for his 5 niece . The gifts include 2 identical Sun and Fun beach dolls , one Elegant eddie dress doll , one GI jose doll , and one tulip troll doll . If th eyoungest niece doesnot want the GI jose doll , in how many different ways can he give the gifts?

OA 48
Required number of ways = Number of ways 5 dolls can be distributed among 5 nieces - Number of ways in which youngest one gets the GI Joe doll

Number of ways 5 dolls can be distributed among 5 nieces = 5! = 120
As there are 2 dolls of same kind, effective number of ways = 120/2 = 60

Number of ways in which youngest one gets the GI Joe doll = Number of ways to distribute other 4 dolls among remaining 4 nieces = 4!/2 = 12 (Division by 2 for the same reason as above)

Required number of ways = 60 - 12 = 48
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by rishab1988 » Sat Nov 27, 2010 8:06 am
Although Rahul has solved this question.I solved in a direct way.

There are 5 dolls and 5 girls. 2 dolls are same.

One of the girls doesn't get a specific doll (G.I Joe doll).

Lets make 5 dashes _ _ _ _ _ (for 5 girls).

We fill the dash with the most restriction first.Since the smallest girl can't be alloted one of the dolls,she can only be alloted one of the 4 remaining dolls.

4 _ _ _ _

Now the second girl can be alloted any of the 4 remaining dolls [ one of the 5 dolls has already been alotted]

4 4 _ _ _

Similarly the third girl can be alloted any of the 3 reaming ones.

4 4 3 _ _

This is same procedure is repeated fot the 2 other girls

4 4 3 2 1.

Now don't forget that the question mentions that 2 of the dolls are identical. So divide by 2!.If 3 dolls were identical ,we would have divided by 3!.

(4*4*3*2*1)/2!=4*4*3=48.

This question was all about logic.I solved it 2 steps.
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