In how many arrangements can a teacher seat 3 girls and 3 bo

Question

In how many arrangements can a teacher seat 3 girls and 3 boys in a row of 6 seats if no two children of the same gender are to be adjacent?

(A) 6
(B) 12
(C) 36
(D) 72
(E) 144

Bunuel · Accepted Answer

BGBGBG or GBGBGB --> 2 ways,

Both 3 B's and 3 G's can be arranged in 3! ways.

Thus toal 2*3!*3!=72.

Answer: D.

EMPOWERgmatRichC · Answer

Hi All,

These types of story-based permutation questions can usually be solved with a basic drawing and some multiplication.

Here, we're told to sit 6 children in a row (3 boys and 3 girls) and that two adjacent seats CANNOT have the same gender. We're asked for the total number of possible ways to seat the 6 children.

So, let's start with 6 chairs:

_ _ _ _ _ _

The first chair can have ANY of the children:

6 _ _ _ _ _

Whether you put a boy or a girl in that first chair, the second chair MUST have the opposite gender - there will be 3 of those children available:

6 3 _ _ _ _

From here, we just have to go "back and forth" with the 2 remaining boys and the 2 remaining girls:

6 3 2 2 1 1

Now, we just multiply the numbers: (6)(3)(2)(2)(1)(1) = 72 possible arrangements

Final Answer:  D

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

Temurkhon · Answer

B G B G B G=3*3*2*2*1*1

OR

G B G B G B=3*3*2*2*1*1

36+36=72

D

gurmukh · Answer

I would request whosoever is doing this sum also try a sum in which you have to find out in how ways 3 boys and 3 girls can be seated so that no two girls are together and two sums are not same and their answers are also different. 
I'll explain why..
In this first 3 boys can be seated in 6 ways
Now there are 4 spaces on the side of boys so girls can be seated in those 4 spaces in 24 ways so answer is 6×24 = 144
The difference between this sum and earlier sum is that in this sum 2 boys can be together which in earlier sum was invalid.

Posted from my mobile device

CEdward · Answer

Permutations with restrictions:

M F M F M F OR 
F M F M F M

For each arrangement, guys and girls can arrange themselves in 3! x 3! = 36 respectively. Since these two arrangements depicted above are mutually exclusive, they cannot occur at the same time. So we have to add.

36 + 36 = 72 ways

Answer is D.

luisdicampo · Answer

Deconstructing the Question  
Seat  3 distinct girls  and  3 distinct boys  in 6 seats.
“No two of the same gender adjacent” forces an  alternating gender pattern .
Key idea: count gender patterns, then permute within each gender.

Step-by-step

Possible gender patterns (must alternate):
 GBGBGB  or  BGBGBG  →  2  patterns.

For each pattern:
Arrange the girls in the 3 girl-slots:
 3!=6 
Arrange the boys in the 3 boy-slots:
 3!=6

Total:
 2\cdot 3!\cdot 3!=2\cdot6\cdot6=72

Answer: 72

In how many arrangements can a teacher seat 3 girls and 3 bo

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GMAT Problem Solving (PS) Questions

In how many arrangements can a teacher seat 3 girls and 3 bo

Prep Toolkit

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