Aptitude Discussion

Q. |
Four boys and three girls stand in queue for an interview. The probability that they stand in alternate positions is: |

✔ A. |
1/35 |

✖ B. |
1/34 |

✖ C. |
1/17 |

✖ D. |
1/68 |

**Solution:**

Option(**A**) is correct

Total number of possible arrangements for 4 boys and 3 girls in a queue $= 7!$

When they occupy alternate position the arrangement would be like:

$B$ $G$ $B$ $G$ $B$ $G$ $B$

Thus, total number of possible arrangements for boys $= (4×3×2)$

Total number of possible arrangements for girls $= (3×2)$

Required probability

\(=\dfrac{(4\times 3\times 2\times 3\times 2)}{7!}\)

\(=\dfrac{1}{35}\)

**Suman**

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The question specifically said that alternative position and choosing position other than BGBGBGB will result in at least 2 boys back to back which is not allowed. Now once this is understood, the 4 boys can shuffle themselves 4! and girls can 3!. so total favourable ways are 4!x3!

and prob is divide it by 7!

**Snehal**

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Girls have 5 places to choose from ....at the start, between boys and a place after the last boy.....They can select the positions in 5C3 ways and then the girls can occupy the three places in 3! ways....therefore girls can be arranged in 5C3* 3! ways....Is my understanding right?

**Pratik**

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What if que starts with a girl

then the boys will repeat like GBGBGBB which is not alternate,

**Niranjana**

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It can also be 1/21

IT can be like GBGBGBGBG like snehal said

then 5C3*3! for girls and 4! for boys