Combinations and Permutations

Question

In how many ways can 12 different books be parceled into 4 packets so that 3 of the packets contain 3 books each and one of them two books atleast, if all the packets have the same destination?

Comments

@Vikas Verma,Finally edited. :)

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@arvin,Could you please work out your steps?I am getting a bit confused while solving. Thanks in advance too. :)

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@devshree done bro :p

Answer 1

See its a tricky question :

1)  we have 12 different books to be send into packets having same address ----> means n distinct objects to be placed into r identical boxes.

 2) now 3 of them contains 3 each and one among them has atleast 2.

means  possible combinations {3,3,3,3} only .

one among them has atleast two books is a dead statement  // it has no reference for question //given only to confuse

so number of ways : (as parcels are identical they will be selected in one way only)

 {3,3,3,3} = 12!/((3!)⁴*4!)           = 15400

#total ways  = 15400 ways.

Comments

@devshree : it could have been done by stirlings method. but since there is a restriction that each among them have 3 elements it will be even more difficult and will be more time consuming.

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@arvin,Then too a big Thanks bro. :). You spared your precious time. :)

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@devshree it's ok bro :) I am also new for combinatorics so it was an exercise for me tooo and I too learner different ways through this question of urs. Thanks a lot