ISI permutation and comb doubt

[SoniaKapoor]

No of ways in which 6 pencils can be distributed btw 2 boys such that each boy gets atleast 1 pencil...


How to solve such problems..

[SoniaKapoor]

Wrong....
the options are 30,60,62,64
ISI 2011 quest 15

[don123]

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[XIPP]

 

[don123]

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[XIPP]

Hi, Baadshah,

The problem should have specified whether the pencils are distinguishable or not. Without that, there is ambiguity, and your interpretation (and answer) are perfectly reasonable.

[don123]

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[SoniaKapoor]

Thankyou so much guys..

[Tania123]

I solved it in the following manner.
Taking the pencils to be distinct, we can first choose 2 pencils out of 6 in 6C2 ways(each boy must have atleast one) These two pencils can be divided among the 2 boys in 2! ways. The remaining 4 pencils can be then divided among the 2 boys in 2^4 ways.
Therefore, total no. of ways= 6C2*2!*(2^4) =480
Where am I gong wrong?

[Dr. Strange]

This method has lots of repetition.
Suppose pencils are marked 1,2,3,4,5,6

Suppose CASE 1:
Initially two pencils 1 and 3 are chosen.
Boy 1 gets pencil 1,Boy 2 gets 3
Finally rest 4 pencils are distributed.
let final allocation is:

Boy 1: 1,4,5
Boy 2:2,3,6

CASE 2:
Initially two pencils 2 and 4
 are chosen.
Boy 1 gets pencil 4,Boy 2 gets 2

Finally rest 4 pencils are distributed.
Let boy 1 gets 1,5
Boy 2 gets 3,6
 final allocation is:

Boy 1: 1,4,5
Boy 2:2,3,6


BOTH CASES ARE IDENTICAL

[Tania123]

But isn't this the general method for calculating the number of ways n distinct objects can be divided into r groups such that each of them get atleast 1?

[Dr. Strange]

No its not the right way.


If n= no. of distinct objects and k=no. of people then expression is:

[shruti malhotra]

u can do it like this also
boy1      boy2    no of ways
1            5         6c1*5c5=6
2             4        6c2*4c4=15
3             3        6c3*3c3=20
4            2         6c4*2c2=15
5            1         6c5*1c1=6
TOTAL WAYS=62
HOPE THIS HELP :)