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by kstv » Fri Jun 11, 2010 12:48 am
1 to 35 has 18 odd 1 - 3 - 5 -..... 33- 35
No one can get 35 or 33 marbles, cos there will not be enough left for choosing the other 4.
If one gets 31 marble the rest four get 1 each.
So there are 16 odd numbers to choose.
My guess is 15 C 4
cos if you choose the 4 odd numbers the fifth one has to be 35 - sum of the other four.
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by blaster » Fri Jun 11, 2010 2:01 am
thanks kstv!
but it would be good to see another explanation also.
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by winnerhere » Fri Jun 11, 2010 10:11 pm
kstv wrote:1 to 35 has 18 odd 1 - 3 - 5 -..... 33- 35
No one can get 35 or 33 marbles, cos there will not be enough left for choosing the other 4.
If one gets 31 marble the rest four get 1 each.
So there are 16 odd numbers to choose.
My guess is 15 C 4
cos if you choose the 4 odd numbers the fifth one has to be 35 - sum of the other four.
Great explanation. Thanks :)
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by winnerhere » Sun Jun 13, 2010 1:41 am
kstv wrote:1 to 35 has 18 odd 1 - 3 - 5 -..... 33- 35
No one can get 35 or 33 marbles, cos there will not be enough left for choosing the other 4.
If one gets 31 marble the rest four get 1 each.
So there are 16 odd numbers to choose.
My guess is 15 C 4
cos if you choose the 4 odd numbers the fifth one has to be 35 - sum of the other four.
kstv,

boy 1 have 16 options [ he selects 1]

boy 2 - 16 options[ he selects 1]

boy 3 - 16 options[he selects 1]

boy 4- 16 options [he selects 1]

boy 5 - he has just one option

so by this logic the asnwer should be 16^4

I hope there should be some other logic behind this.

It is like a+b+c+d+e = 35

where a,b,c,d,e can only take odd numbers.
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by albatross86 » Sun Jun 13, 2010 4:16 am
I used a different approach and got a different answer.

We have to make sure each boy gets an odd number of marbles.

First, give each boy 1 marble. This leaves 30 marbles.

We can now distribute the remaining 30 marbles in 15 groups of 2 (Since they are identical). This ensures that since we are now giving an even number of marbles to each boy, this added to the original 1 marble gives us an odd number for each boy.

So the problem becomes one of distributing 15 objects to 5 boys.

This the situation of combinations of putting n objects in r boxes with repetition / empty boxes allowed, for which the formula is C (n + r -1 , r - 1) [This can be derived, but just remember it for the GMAT imo]

= C ( 15 + 5 -1, 5 - 1 )
= C (19, 4)

Pick B.
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by winnerhere » Sun Jun 13, 2010 8:32 am
albatross86 wrote:I used a different approach and got a different answer.

We have to make sure each boy gets an odd number of marbles.

First, give each boy 1 marble. This leaves 30 marbles.

We can now distribute the remaining 30 marbles in 15 groups of 2 (Since they are identical). This ensures that since we are now giving an even number of marbles to each boy, this added to the original 1 marble gives us an odd number for each boy.

So the problem becomes one of distributing 15 objects to 5 boys.

This the situation of combinations of putting n objects in r boxes with repetition / empty boxes allowed, for which the formula is C (n + r -1 , r - 1) [This can be derived, but just remember it for the GMAT imo]

= C ( 15 + 5 -1, 5 - 1 )
= C (19, 4)

Pick B.
awesome explanation - its right answer :)

the remaining 30 marbles are divided into 15 objects[1 object = 2 balls]

now we have to distribute 15 objects to 5 boys = 19C4 :)

Thanks albatross
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