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Three gnomes and three elves sit down in a row of six chairs

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by AAPL » Wed Jun 12, 2019 4:32 am

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Manhattan Prep

Three gnomes and three elves sit down in a row of six chairs. If no gnome will sit next to another gnome and no elf will sit next to another elf, in how many different ways can the elves and gnomes sit?

A. 18
B. 36
C. 48
D. 72
E. 96

OA D
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by Ian Stewart » Wed Jun 12, 2019 5:59 am
If there's a gnome in the first chair, the arrangement must alternate: GEGEGE. We'll have 3 choices for the first gnome, 2 for the second, and 1 for the third, and the same for the elves, for 3! * 3! = 36 arrangements in total. But we can also have an elf in the first chair, and the arrangement could be EGEGEG, which will yield a further 36 arrangements. So we have 72 in total.
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by Scott@TargetTestPrep » Fri Jun 14, 2019 2:39 pm
AAPL wrote:Manhattan Prep

Three gnomes and three elves sit down in a row of six chairs. If no gnome will sit next to another gnome and no elf will sit next to another elf, in how many different ways can the elves and gnomes sit?

A. 18
B. 36
C. 48
D. 72
E. 96

OA D
We can see that one seating arrangement of gnomes (G) and elves (E) can be:

GEGEGE

The first G and E each has 3 choices; the second G and E each has 2 choices and the last G and E each has 1 choice. Thus the number of ways to have the "GEGEGE" seating arrangement is:

3 x 3 x 2 x 2 x 1 x 1 = 36

However, the seating arrangement can also be EGEGEG, and there will also be 36 such arrangements. Thus, the total number of seating arrangements is 36 + 36 = 72.

Answer: D

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