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5 boys and 5 girls randomly select seats around a circular t

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by Max@Math Revolution » Fri Jun 21, 2019 12:55 am

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[GMAT math practice question]

5 boys and 5 girls randomly select seats around a circular table that seats 10. What is the probability that two girls will sit next to one another?

A. 11/24
B. 23/24
C. 23/48
D. 47/48
E. 125/126
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by Max@Math Revolution » Sun Jun 23, 2019 5:15 pm
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The easiest way to solve this problem is to find the number of arrangements satisfying the complementary condition that boys and girls are arranged alternately around the table, and subtract this from the total number of arrangements of the boys and girls.

The number of arrangements of n people in a circle is (n-1)!.
So, the total number of arrangements of 10 people is (10-1)! = 9!
The number of complementary arrangements is (5-1)!*5! = 4!*5!
Thus, the required probability is 1 - [(4!)(5!)]/(9!) = 1 - [(1*2*3*4)/(6*7*8*9)] = 1 - 1/126 = 125/126

Therefore, E is the answer.
Answer: E
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by deloitte247 » Sat Jun 29, 2019 10:37 am
Circular arrangement of 5 boys and 5 girls=
$$\left(n-1\right)!=\left[\left(5+5\right)-1\right]!=\left[10-1\right]!=9!$$
Where 2/more girls do not sit together =
Ways in which boys can be seated * ways in which boys can be seated.
with the boys going first, they can be seated in 4! ways and the girls can be seated in 5! ways.
Total number of ways 10 people could be seated round the circular table = 9!
The probability =
$$=\frac{9!-4!\cdot5!}{9!}$$
$$=\frac{\left(9\cdot8\cdot7\cdot---\cdot1\right)-\left(4\cdot3\cdot2\cdot1\right)\cdot\left(5\cdot4\cdot3\cdot2\cdot1\right)}{\left(9\cdot8\cdot7\cdot-------\cdot1\right)}$$
$$=\frac{\left(362880\right)-\left(24\cdot120\right)}{362880}$$
$$=\frac{\left(360000\right)}{362880}=\frac{125}{126}$$

$$answer\ is\ Option\ E$$
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