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08 Jun 2022, 01:30
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A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
45%
(medium)
Question Stats:
68% (01:52) correct
32%
(02:29)
wrong
based on 251
sessions
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In a particular 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
Are You Up For the Challenge: 700 Level Questions
A. 105
B. 210
C. 315
D. 420
E. 630
Are You Up For the Challenge: 700 Level Questions
Updated on: 08 Jun 2022, 05:48
Originally posted by GMATGuruNY on 08 Jun 2022, 04:00.
Last edited by GMATGuruNY on 08 Jun 2022, 05:48, edited 1 time in total.
Last edited by GMATGuruNY on 08 Jun 2022, 05:48, edited 1 time in total.
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Bunuel
The 3 red gumballs selected at the end are irrelevant, since they can be arranged only 1 way:
RRR
Thus, the question stem can be rephrased as follows:
Before the 3 red gumballs are selected at the end, in how many different ways can the other 7 gumballs be selected?
Number of ways to arrange 7 distinct elements = 7!
But the elements here are not all distinct; some are IDENTICAL.
When an arrangement includes identical elements, we must divide by the number of ways each set of identical elements can be arranged.
The reason:
When the identical elements swap positions, the arrangement doesn't change, reducing the total number of unique arrangements.
Here, we must divide by the number of ways to arrange the 4 identical blue gumballs (4!) and the number of ways to arrange the 2 identical green gumballs (2!):
\(\frac{7!}{4!2! } = 105\)
08 Jun 2022, 04:05
Kudos
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Since 3 red gumballs have to be dispatched last, the other 7 balls should be dispatched first in any order.
The 4 blue, 2 green and 1 yellow ball can arrange themselves in 7!/4!*2! ways
7*6*5/2 = 105.
Thus the correct option will be A.
The 4 blue, 2 green and 1 yellow ball can arrange themselves in 7!/4!*2! ways
7*6*5/2 = 105.
Thus the correct option will be A.
