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There are 6 children at a family reunion, 3 boys and 3 girls

tagged by: BTGmoderatorLU

This topic has 3 expert replies and 1 member reply

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There are 6 children at a family reunion, 3 boys and 3 girls

There are 6 children at a family reunion, 3 boys, and 3 girls. They will be lined up single-file for a photo, alternating genders. How many arrangements of the children are possible for this photo?

A. 54
B. 72
C. 18
D. 81
E. 36

The OA is B.

I'm confused by this PS question. Experts, any suggestion? I always have troubles with the combination questions.. Thanks in advance.

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We know that we *must* alternate boys and girls. This means we have two options for the order of boys and girls:

B--G--B--G--B--G
or
G--B--G--B--G--B

Within each of these two options, the boys can be arranged in any order and the girls can be arranged in any order. So we can think about this problem as:

2 order options * possible arrangements of 3 boys * possible arrangements of 3 girls

There are 3 boys and 3 girls. This means that the total number of arrangements for each is 3! - 3*2*1 = 6. So our equation is
2 order options * 6 arrangements of 3 boys * 6 arrangements of 3 girls
2 * 6 * 6 = 72

So there are 72 total arrangements of children available for the photo.

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Hi LUANDATO,

We're told that 3 boys and 3 girls will be lined up single-file for a photo, alternating genders. We're asked for the number of arrangements of the children that are possible for this photo. This question is a fairly standard Permutation question with one small 'twist.'

Since the first child in line can be EITHER a boy or a girl, there are 6 options for that 1st spot. Once you put someone there though, the 2nd spot must be the OTHER gender, so...
there are 3 spots for the 2nd spot.

From there, the genders are boy-girl-boy-girl or girl-boy-girl-boy (depending who was in the 1st spot...
there are 2 spots for the 3rd spot.
there are 2 spots for the 4th spot.
there are 1 spots for the 5th spot.
there are 1 spots for the 6th spot.

Total arrangements = (6)(3)(2)(2)(1)(1) = 72

GMAT assassins aren't born, they're made,
Rich

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Contact Rich at Rich.C@empowergmat.com

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LUANDATO wrote:
There are 6 children at a family reunion, 3 boys, and 3 girls. They will be lined up single-file for a photo, alternating genders. How many arrangements of the children are possible for this photo?

A. 54
B. 72
C. 18
D. 81
E. 36
We can have the following arrangements:

BGBGBG or GBGBGB

Since the number of people in the arrangement is the same, the number of arrangements for each is the same. Let’s tackle the first arrangement.

For the first position, we have 3 possible boys to choose from. For the second position, we have 3 girls to choose from. For the third position, we have 2 boys, and for the fourth position we have 2 girls, and for the fifth, there is 1 boy, and for the sixth, there is 1 girl. Thus:
3 x 3 x 2 x 2 x 1 x 1 = 36 ways

So the total number of ways is 36 + 36 = 72 ways.

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Jeffrey Miller Head of GMAT Instruction

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Firstly, let us arrange the boys in a way that there is a gap that will be filled by a girl in between each boy.
There are 3 boys to arrange in 3 positions. Therefore the number of permutation possible= $$3P_2$$ = 6ways
Next, we fill up the remaining 3 gaps with 3 girls .
Number of possible permutation = $$3P_2$$ = 6ways
We have permuted the possibilities in this way B-G-B-G-B-G
But we can also have the arrangement in this way: G-B-G-B-G-B
This means that for every arrangement obtained earlier, there are 2 possible permutations.
Therefore the total number of arrangement possible = 2 * [6*6] =72 possible arrangements

we can confidently say option B is correct

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