In a particular gumball machine, there are 4 identical blue

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AAPL: Moderator; Posts: 2632; Joined: Sun Oct 29, 2017 2:08 pmFollowed by: 2 members

by AAPL » Tue Jun 26, 2018 5:46 am

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In a certain gumball machine, there are 4 identical blue gumballs, 3 identical red gumballs, 2 identical green gumballs, and 1 yellow gumball. In how many different ways can the gumballs be dispensed, 1 at a time, if the 3 red gumballs are dispensed last?

A. 105
B. 210
C. 315
D. 420
E. 630

The OA is A.

Is there a strategic approach to this question? Can anyone help, please? Thanks!

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Source: Beat The GMAT — Problem Solving | Tue Jun 26, 2018 5:46 am

regor60: Master | Next Rank: 500 Posts; Posts: 416; Joined: Thu Oct 15, 2009 11:52 am; Thanked: 27 times

by regor60 » Tue Jun 26, 2018 7:21 am

AAPL wrote:In a certain gumball machine, there are 4 identical blue gumballs, 3 identical red gumballs, 2 identical green gumballs, and 1 yellow gumball. In how many different ways can the gumballs be dispensed, 1 at a time, if the 3 red gumballs are dispensed last?

A. 105
B. 210
C. 315
D. 420
E. 630

The OA is A.

Is there a strategic approach to this question? Can anyone help, please? Thanks!

With the 3 red gumballs being dispensed at the end you the number of combinations is due solely to the blue, green and yellow gumballs.

So, with 7 slots available, there 7! different ways to arrange the gumballs, treating them as distinct.

But because there are 4 indistinguishable blues and 2 greens, you need to divide 7! by 4!x2!, or 7!/4!x2! in order to eliminate the duplicate combinations, so (7x6x5x4x3x2x1)/((4x3x2x1)(2x1)) = A, 105

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