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Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
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5 boys and 5 girls randomly select seats around a circular table that  [#permalink]

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Difficulty:   55% (hard)

Question Stats: 65% (03:03) correct 35% (02:24) wrong based on 23 sessions

### HideShow timer Statistics [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?

$$A. \frac{11}{24}$$

$$B. \frac{23}{24}$$

$$C. \frac{23}{48}$$

$$D. \frac{47}{48}$$

$$E. \frac{125}{126}$$

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Re: 5 boys and 5 girls randomly select seats around a circular table that  [#permalink]

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What we will do to find the number of ways two/more girls will sit together is to find the total number of arrangements of 5 boys and 5 girls in a circle and subtract those arrangemments where two/more girls DO NOT sit together. To do that, we must alternate them like this BGBGBGBGBG. We will seat the boys first.They can be seated in 4! ways. Then the girls will be seated in 5! ways. Now the total number of ways 10 people can be arranged in a circle is 9!The probability will then be
(9!-4!*5!/9! which would give you the value in E
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Re: 5 boys and 5 girls randomly select seats around a circular table that  [#permalink]

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1
for circular arragement total ways to sit will be (n-1)! ; (10-1)! ; 9! WAYS
for boys let the first position be fixed so we have 4! ways to arrange sitting
for girls not sitting together they can be made to sit in 5! ways ; note that this is circular arrangement so 5! *2 wont be possible
total ways that girls sit next to each other
1-4!*5!/9! ; 9!-4!*5!/9! = ~ 0.99
IMO E

MathRevolution wrote:
[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?

$$A. \frac{11}{24}$$

$$B. \frac{23}{24}$$

$$C. \frac{23}{48}$$

$$D. \frac{47}{48}$$

$$E. \frac{125}{126}$$

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Math Revolution GMAT Instructor V
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5 boys and 5 girls randomly select seats around a circular table that  [#permalink]

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1
=>

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 $$1 – \frac{[(4!)(5!)]}{(9!)} = 1 – \frac{[(1*2*3*4)}{(6*7*8*9)]} = 1 – \frac{1}{126} = \frac{125}{126}$$

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Re: 5 boys and 5 girls randomly select seats around a circular table that  [#permalink]

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HI Bunuel, EgmatQuantExpert, amanvermagmat, GMATGuruNY, GMATPrepNow, pushpitkc, IanStewart

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 $$(10-1)! = 9!$$

How to find the number of complementary arrangements?

I'm a confused bit in the below steps...

Let boys occupy alternate seat i,e 5! as it is circular arrangement hence (5-1)! = 4!

The remaining alternate seat will be occupied by 5 girls 5!? Here why it can't be (5-1)! = 4!
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Re: 5 boys and 5 girls randomly select seats around a circular table that  [#permalink]

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NandishSS wrote:
I'm a confused bit in the below steps...

Yes, I'd find that a confusing way to solve the problem too.

So that we have no girls sitting together, we just need boys and girls to alternate. So we either need a seating arrangement like BGBGBG... or like GBGBGB... Now it doesn't matter if the seats are in a circle or in a triangle or just in a line on a stage - there's no reason to use circular permutation formulas. If we imagine a line of chairs on a stage, there will be 2 arrangements which alternate between boys and girls (or girls and boys). The total number of arrangements possible is 10C5 = (10)(9)(8)(7)(6)/5! = 252, because we just need to choose 5 out of the 10 seats for the boys. So the probability our arrangement alternates is 2/252 = 1/126, and the probability it doesn't is 125/126.
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5 boys and 5 girls randomly select seats around a circular table that  [#permalink]

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We don't have any reference point until boys sit, hence the [$$B_1 B_2 B_3 B_4 B_5$$] pattern count as same for any rotation.
Hence total seating combination for boys= 5!/5=4!

Now, once you fixed the seat for the boys, you actually have the reference point now. Hence [$$G_1 G_2 G_3 G_4 G_5$$] gonna count as different for every rotation.
Hence total seating combinations of girls =5!

The top 2 cases count as same but bottom two cases count as different in the following figure. But as IanStewart said that the shape of the seating pattern doesn't matter, if you have to find the probability. So you can treat this case as linear arrangement. But if you have to find the number of combinations, it will matter.

NandishSS wrote:
HI Bunuel, EgmatQuantExpert, amanvermagmat, GMATGuruNY, GMATPrepNow, pushpitkc, IanStewart

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 $$(10-1)! = 9!$$

How to find the number of complementary arrangements?

I'm a confused bit in the below steps...

Let boys occupy alternate seat i,e 5! as it is circular arrangement hence (5-1)! = 4!

The remaining alternate seat will be occupied by 5 girls 5!? Here why it can't be (5-1)! = 4!

Attachments f.png [ 20.74 KiB | Viewed 349 times ]

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GMAT 1: 800 Q51 V49 GRE 1: Q170 V170 Re: 5 boys and 5 girls randomly select seats around a circular table that  [#permalink]

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Hi NandishSS,

This question can be approached in a number of different ways (depending on how you choose to do the math involved). It's far easier to calculate the probability that NO girls will sit next to one another, so I'm going to calculate that and then subtract that fraction from the number 1; this will give us the probability that two (or more) girls will end up sitting next to one another.

To start, any of the 10 chairs can be the 'first chair' - and any of the 10 people can sit in it. Once someone is placed in that seat, we want the next seat have someone of the opposite sex (and there are 5 options for that 'second chair'). From here, the pattern is now set for the other 8 chairs (it's either BGBGBGBG or GBGBGBGB, depending on who sits in the 'first chair') - and we'll end up with 4 possible kids for the 'third chair', 4 for the 'fourth chair', 3 for the 'fifth chair', 3 for the 'sixth chair', etc.

This means that there are (10)(5)(4)(4)(3)(3)(2)(2)(1)(1) possible ways to arrange the boys and girls in every-other-seat. Do NOT multiply that out yet.

The total possible ways to place the 10 people in the 10 seats is 10!, which I'm going to expand-out in a moment. We now have our probability:

(10)(5)(4)(4)(3)(3)(2)(2)(1)(1) / (10)(9)(8)(7)(6)(5)(4)(3)(2)(1)

You can 'cancel out' a LOT of the numbers in the numerator and denominator. For example, the 10s 'cancel out', you can take the two 3s from the numerator and 'cancel out' the 9 in the denominator, etc. By doing that, I was left with:

(1) / (7)(6)(3) = 1/126

Thus, since there's a 1/126 probability of having NO girls sit next to one another, there's a 1 - (1/126) = 125/126 probability of having 2 (or more) girls sit next to each other.

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Re: 5 boys and 5 girls randomly select seats around a circular table that  [#permalink]

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MathRevolution wrote:
[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?

$$A. \frac{11}{24}$$

$$B. \frac{23}{24}$$

$$C. \frac{23}{48}$$

$$D. \frac{47}{48}$$

$$E. \frac{125}{126}$$

The only way that two girls will not sit next to one another is if the boys and girls sit alternately next to one another.

10!/(5! x 5!) = (10 x 9 x 8 x 7 x 6)/(5 x 4 x 3 x 2) = 3 x 2 x 7 x 6 = 252

The number of ways BGBGBGBGBG can sit is:

5 x 5 x 4 x 4 x 3 x 3 x 2 x 2 x 1 x 1 = (5!)^2

The number of ways GBGBGBGBGB can sit is also (5!)^2.

Therefore, the probability is:

1 - 2(5!)^2/10! = 1 - 1/126 = 125/126

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