# Easy Permutation-Combination Solved QuestionAptitude 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.

## (11) Comment(s)

V J
()

$Explanation:$

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!= $384$

Satish
()

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
()

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
()

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

Martin
()

Is this correct?

I've enumerated 768 arrangements.

Abi
()

coudnt understand the question and its explanation!

Shaswat Khamari
()

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

Shamsa Kanwal
()

there should be 8 people shouldn't

Sagar
()

Even after that I think the answer is wrong..

Chirag Khimani
()

ya I think it is a mistake...

Gail Mitchell
()

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