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Amanda goes to the toy store to buy 1 ball and 3 different board games

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Amanda goes to the toy store to buy 1 ball and 3 different board games  [#permalink]

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New post 27 Dec 2015, 08:48
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Question Stats:

93% (00:46) correct 7% (00:55) wrong based on 127 sessions

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Amanda goes to the toy store to buy 1 ball and 3 different board games. If the toy store is stocked with 3 types of balls and 6 types of board games, how many different selections of the 4 items can Amanda make?

A. 9
B. 12
C. 14
D. 15
E. 60

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Re: Amanda goes to the toy store to buy 1 ball and 3 different board games  [#permalink]

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New post 30 Dec 2015, 18:10
Amanda goes to the toy store to buy 1 ball and 3 different board games. If the toy store is stocked with 3 types of balls and 6 types of board games, how many different selections of the 4 items can Amanda make?


3! / 1!2! * 6! / 3!3!
=3*20=60

E. 60
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Re: Amanda goes to the toy store to buy 1 ball and 3 different board games  [#permalink]

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New post 02 Jan 2016, 19:23
Hi All,

The answer choices to this question provide an interesting 'shortcut' that you can use to avoid some of the math involved.

We're told that Amanda is going to buy 3 different board games (from a total of 6 different possible games). Since the order of the games does NOT matter, we're dealing with a Combination Formula calculation:

Combinations = N!/K!(N-K)! where N is the total number of items and K is the size of the subgroup.

N = 6 and K = 3

6!/3!(6-3)! = (6)(5)(4)(3)(2)(1)/(3)(2)(1)(3)(2)(1) = (6)(5)(4)/(3)(2)(1) = 20 different combinations of 3 games.

Since we already have 20 different options, including a ball will only INCREASE the number of possibilities. There's only one answer that fits...

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Re: Amanda goes to the toy store to buy 1 ball and 3 different board games  [#permalink]

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New post 02 Jan 2016, 23:33
Bunuel wrote:
Amanda goes to the toy store to buy 1 ball and 3 different board games. If the toy store is stocked with 3 types of balls and 6 types of board games, how many different selections of the 4 items can Amanda make?

A. 9
B. 12
C. 14
D. 15
E. 60


This is a simple question that tests your concepts about combinations.

We need to select 1 ball out of 3 available and 3 board games out of 6 available.
This can be done in 3C1*6C3 ways = 3*\(\frac{{6!}}{{3!*3!}}\) = 3*20 = 60

Option E
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Re: Amanda goes to the toy store to buy 1 ball and 3 different board games  [#permalink]

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New post 03 Jan 2016, 03:24
Pretty Straight.

Choose 1 ball from 3 - 3C1 = 3/1 = 3
Choose 3 board games from 6 - 6C3 = (6*5*4)/(1*2*3)= 20

Total number of ways the selection can happen is = 3 * 20 = 60
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Re: Amanda goes to the toy store to buy 1 ball and 3 different board games  [#permalink]

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New post 02 Apr 2016, 02:19
3C1*6C3=3*20 = 60

Correct answer - E
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Re: Amanda goes to the toy store to buy 1 ball and 3 different board games  [#permalink]

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New post 26 Oct 2017, 05:44
I do not get logic. The explaination in the Kaplan book doesn't tell clearly why you multiply 3*20 and why you use the combination formula.
can someone explain why you multiply 3*20 and why you use combination insteat of permutation?
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Re: Amanda goes to the toy store to buy 1 ball and 3 different board games  [#permalink]

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New post 30 Oct 2017, 13:52
Bunuel wrote:
Amanda goes to the toy store to buy 1 ball and 3 different board games. If the toy store is stocked with 3 types of balls and 6 types of board games, how many different selections of the 4 items can Amanda make?

A. 9
B. 12
C. 14
D. 15
E. 60


Amanda can select a ball in 3C1 = 3 ways and she can selected a board game in 6C3 = 6!/[3!(6-3)!] = (6 x 5 x 4)/3! = (6 x 5 x 4)/(3 x 2 x 1) = 20. So, the total number of ways is 3 x 20 = 60.

Answer: E
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Re: Amanda goes to the toy store to buy 1 ball and 3 different board games &nbs [#permalink] 30 Oct 2017, 13:52
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