Aptitude Discussion

Q. |
$12$ chairs are arranged in a row and are numbered $1$ to $12$. $4$ men have to be seated in these chairs so that the chairs numbered $1$ ans$8$ should be occupied and no two men occupy adjacent chairs. Find the number of ways the task can be done. |

✖ A. |
$360$ |

✔ B. |
$384$ |

✖ C. |
$432$ |

✖ D. |
$470$ |

**Solution:**

Option(**B**) is correct

Given there are $12$ numbered chairs, such that chairs numbered $1$ and $8$ should be occupied.

$\_X\_ , 2,3,4,5,6,7,\_X\_,9,10,11,12.$

The various combinations of chairs that ensure that no two men are sitting together are listed.

$(1,3,5,\_\_)$. The fourth chair can be $5,6,10,11$ or $12$, hence $5$ ways.

$(1,4,8,\_\_)$, The fourth chair can be $6,10,11$ or $12$ hence $4$ ways.

$(1,5,8,\_\_)$, the fourth chair can be $10,11$ or $12$ hence $3$ ways.

$(1,6,8,\_\_)$, the fourth chair can be $10,11$ or $12$ hence $3$ ways.

$(1,8,10,12)$ is also one of the combinations.

Hence, $16$ such combinations exist.

In the case of each these combinations, we can make the four men inter arrange in $4!$ ways.

Hence, the required result,

$= 16× 4! = \textbf{384}$

**Edit:** After input from **Gail Mitchell** question hase been updated from '$1$ to $8$' to '$1$ and $8$'.

**Edit:** for an alternative solution, check comment from **Raj.**

**V J**

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**Satish**

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Please frame questions correctly, otherwise, it will kill user's confidence ,interest. ie; Exactness, appropriate words. increase difficulty in L2, L3 levels.Let users able to understand easily in Basic level

**Raj**

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If position 1 and 8 are occupied then when

- Both the remaining men sitting left of 8......then the following positions are possible for the two men (3,5);(3,6) and (4,6).....ie 3 arrangement possible
- Both the remaining men sitting right of 8.... only position (10,12) is possible for them to occupy.....ie 1 arrangement possible.
- One remaining man sitting left of 8 and the other sit right of 8 in this case left man can occupy any of (3,4,5,6) position and right man can occupy any of (10,11,12) position.....ie $4*3=12$ arrangement is possible.

Thus, in total $3+1+12=16$ arrangement are possible.

and 4 men can be arranged in 4! ways, therefore, answer $=16*4!$

**Sushma Saroj**

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I also didn't understand the question and it's answer.plz help me

**Martin**

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Is this correct?

I've enumerated 768 arrangements.

**Abi**

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coudnt understand the question and its explanation!

**Shaswat Khamari**

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It should be "chairs 1 and 8 must be occupied."

**Shamsa Kanwal**

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there should be 8 people shouldn't

**Sagar**

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Even after that I think the answer is wrong..

**Chirag Khimani**

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ya I think it is a mistake...

**Gail Mitchell**

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Shouldn't the wording be "the chairs 1 AND 8" must be occupied?

$Explanation:$

Two chairs are definitely occupied, i.e, 1 and 8. Hence, we just need to find the position of next 2 men.

Now, 2,7 and 9 can never be occupied, since they are supposed to be seated alternatively.

Thus, the combinations go as below:

if 3rd man occupies chair #3, 4th man can occupy chairs 5,6,10,11,12=5 possibilities

if 3rd man occupies chair #4, 4th man can occupy chairs 6,10,11,12=4 possibilities

if 3rd man occupies chair #5, 4th man can occupy chairs 10,11,12=3 possibilities

if 3rd man occupies chair #6, 4th man can occupy chairs 10,11,12=3

if 3rd man occupies chair #10, 4th man can occupy chair #12= 1

Hence, total number of possibilities are: 5+4+3+3+1=$16$

Also, those 4 men can interchange their places in $4!$ ways

Hence, total number of ways= 16 x 4!= $384$