In how many ways can 6 boys be allotted into 5 rooms such that no room

Question

In how many ways can 6 boys be allotted into 5 rooms such that no room is empty and all 6 boys are accommodated?

A. 27
B. 720
C. 840
D. 900
E. 1800

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chetan2u · Answer

So, first we choose two boys who would stay together =>6C2
Now we have rooms and 5 different groups, 2, 1, 1, 1, 1....These 5 can be accommodated in 5! ways.

Total ways =  6C2*5!=\frac{6*5}{2}*120=15*120=1800

E

gmatexam439 · Answer

For 6 boys to be placed in 5 rooms, 1 room will have 2 boys.

choose 2 out of 6 boys: 6C2 = 15

5 rooms can be permuted in 5! ways

Therefore, 5*4*3*2*1*15 = 1800

Correct answer = E

danbrown9445 · Answer

For no room to be empty, 2 boys will be allotted to 1 room and the remaining 4 rooms will have 1 boys each.
Ways to allocate 2 boys to 1st room = 6C2 = 15
Ways to allocate 1 boy to 2nd room = 4C1
Ways to allocate 1 boy to 3rd room = 3C1
Ways to allocate 1 boy to 4th room = 2C1
Ways to allocate 1 boy to 5th room = 1C1

Total ways = 15*4*3*2 = 360
Since any one of the 5 rooms can hold 2 boys,multiply result by 5 = 360*5 = 1800

bumpbot · Answer

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In how many ways can 6 boys be allotted into 5 rooms such that no room

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In how many ways can 6 boys be allotted into 5 rooms such that no room

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