Counting Word Problem

[GaiaAir]

I am studying for my GREs and I came across a Counting Problem in the Manhattan GMAT Word Translations Guide that I am having difficulty solving.

The question is:

Gordon buys 5 dolls for his 5 nieces. The gifts include two identical Sun-and-Fun beach dolls, one Elegant Eddie dress-up doll, one G.I. Josie army doll, and one Tulip Troll doll. If the youngest niece does not want the G.I. Josie doll, in how many different ways can he give the gifts?

My initial approach was to do the # of ways to ignore the rule: 2 x 1 [sun-fun dolls] x 3 x 2 x 1 = 12. Number of ways to break the rule: 2 x 1 [sun-fun dolls] x 2 x 1 x 1 = 4. Subtracting the # of ways to break the rule from # way to ignore the rule = 12 - 4 = 8. This is not correct though.

[chrismagoosh]

Hi Gaia Air,

A good way of thinking of this problem is to set up your usual permutation problem. We are arranging four dolls: 4 x 3 x 2 x 1 = 24. In half of these instances, the sun-and-fun doll is redundant. Meaning D1 D2 D3 SF1 SF2 is the same as D1 D2 D3 SF 2 SF 1. So we divide 24/2 = 12.

It seems like when you set up the part about ignoring the rule, you should have had 4 x 3 x 2 x 1, not 2 x 1 x 3 x 2 x 1.

Hope that helps .

[GaiaAir]

Hi Chris,

According to the solution, the answer is 48. Can you please clarify what the answer is?

Thanks!

[chrismagoosh]

Hmmm...it looks like I totally misinterpreted the question .

And it's more complicated than I had thought. The best way to solve it is to figure out the total possibilities given the different possible the dolls the youngest niece is holding.

If she is hold the SNF doll, then the other dolls can be distributed 4! = 24 amongst the other four nieces. If she is holding a EE doll, the other dolls can be broken up 4!/2 (remember we want to divide by 2 to account for the redundancy of the SNF doll). This give us 12. The same math applies when we assume the youngest niece has TT doll: 12 possibilities result. Also remember that we do not want to account for the second time the youngest niece has a SNF doll.

That gives us a total of 12 + 12 + 24 = 48.

Sorry for any confusion .

[Fahad Ali]

Hmmm...it looks like I totally misinterpreted the question .

And it's more complicated than I had thought. The best way to solve it is to figure out the total possibilities given the different possible the dolls the youngest niece is holding.

If she is hold the SNF doll, then the other dolls can be distributed 4! = 24 amongst the other four nieces. If she is holding a EE doll, the other dolls can be broken up 4!/2 (remember we want to divide by 2 to account for the redundancy of the SNF doll). This give us 12. The same math applies when we assume the youngest niece has TT doll: 12 possibilities result. Also remember that we do not want to account for the second time the youngest niece has a SNF doll.

That gives us a total of 12 + 12 + 24 = 48.

Sorry for any confusion .

chrishagoosh can you explain the logic behind division of 2 for removal of redundancy.

[Fahad Ali]

The way i approached the question. 5 kids and 5 dolls. One kid doesn't want a specific doll. So no of ways to give gift to that kid = 4. Now we are left with 4 dolls and 4 kids. So we have 4! ways to distribute the gifts. Why cant we multiply 4 with 4! to get the answer?

[Fahad Ali]

ok now I got it.

Neices ------------ 1 2 3 4 5
No of Dolls ------ S S E G T

If the youngest gets an S, then the rest of the four has 24 (4!) options.
If the youngest gets the E, then the rest of the four has 12 (4!/2!) options. Divided by two in order to account for the repeating S. Just like permutation of repeating letter in a word like SCHOOL.
If the youngest gets the T, then the rest of the four has 12 (4!/2!) options.

total options = 24+12+12 = 48