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how to solve this type of questions..............

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Master | Next Rank: 500 Posts
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by 4GMAT_Mumbai » Mon Aug 09, 2010 9:04 pm
Hi,

Interesting Question ... Thanks !

Case 1: Two people get 2 coins each and 1 guy gets the rest:

2+2+9
2+9+2
9+2+2

3 ways.

Case 2: One person gets 2 coins while the others get more than 2.

Let A get 2 coins while B and C share the rest.

B and C get their basic 2 coins first.

What is the balance? 13 - 2 - 2 - 2 = 7.

These 7 coins are distributed between B and C.

Place them on a table. Let '1' represent a coin.

1 1 1 1 1 1 1

The question boils down to the 'number of ways of placing a + sign in one of the 6 empty positions between these 7 coins. There are 6 empty slots. Hence, 6C1 ways = 6 ways

6 ways for B and C to get additional coins
Similarly, 6 ways for A and B to get additional coins
Similarly, 6 ways for A and C to get additional coins.

Hence, 18 ways for case 2.

Case 3: All 3 get additional coins

Each of A, B and C get their basic 2 coins.

Rest ... Place them as '1's on a table

1 1 1 1 1 1 1

It boils down to placing two '+' signs in the 6 empty slots between these coins. 6C2 = 15.

Hence, 15 ways.

Overall = Case 1 + Case 2 + Case 3

= 3 + 18 + 15 = 36 ways.

Look forward to know if answer is correct and flaws in thinking ...

Thanks.

PS: Doubt if this could be a GMAT question.
Naveenan Ramachandran
4GMAT, Dadar(W) & Ghatkopar(W), Mumbai
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by clock60 » Mon Aug 09, 2010 10:19 pm
also got 36
at first give everybody 2 coins. and left with 13-2*3=7 coins
with no any restrictions 7 coins can be distributed among 3 persons
in 9!/(7!*2!)=36 ways
*******||.
or
***|*|***
or
*****||**
or.....
total number of ways=36
Last edited by clock60 on Mon Aug 09, 2010 11:29 pm, edited 1 time in total.
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by aarati » Mon Aug 09, 2010 11:04 pm
clock60 wrote:also got 36
at first give everybody 2 coins. and left with 13-2*6=7 coins
with no any restrictions 7 coins can be distributed among 3 persons
in 9!/(7!*2!)=36 ways
*******||.
or
***|*|***
or
*****||**
or.....
total number of ways=36
thank you very much........ i understood logic.......i think formula is
total number of gold coins!/at least 2 gold coins !*reaming coins!
correct me if i am wrong.....give me a correct formula.....
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by shekhar.kataria » Tue Aug 10, 2010 12:39 am
clock60 wrote:also got 36
at first give everybody 2 coins. and left with 13-2*3=7 coins
with no any restrictions 7 coins can be distributed among 3 persons
in 9!/(7!*2!)=36 ways
*******||.
or
***|*|***
or
*****||**
or.....
total number of ways=36
How can 7 coins can be distributed among 3 persons in 9!/(7!*2!) ways, Please explain, confused here

any other way approaching this question?
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by likithae » Wed Aug 11, 2010 12:19 am
clock60 wrote:also got 36
at first give everybody 2 coins. and left with 13-2*3=7 coins
with no any restrictions 7 coins can be distributed among 3 persons
in 9!/(7!*2!)=36 ways
*******||.
or
***|*|***
or
*****||**
or.....
total number of ways=36

thanks ,but how 7 coins can be distributed among 3 persons...explain me
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by mallyaprs » Wed Aug 11, 2010 11:00 am
Great solution above, I want to elaborate little on it

13 gold coins are similar, so think of it as dividing similar objects among different people

now you can consider, a,b and c as three people

then

a+b+c= 13

the condition is at least 2 , ie give assign a,b and c 2

then you are left with

a+b+c= 7

now the answer is 7+2 C 2 , where 2 represents the add symbol.

This also represents the number of solutions for the equation a+b+c= 13
This also represents the number of terms in the expansion (a+b+c)^13

hope this helps
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by clock60 » Wed Aug 11, 2010 11:28 am
aarati wrote:
clock60 wrote:also got 36
at first give everybody 2 coins. and left with 13-2*6=7 coins
with no any restrictions 7 coins can be distributed among 3 persons
in 9!/(7!*2!)=36 ways
*******||.
or
***|*|***
or
*****||**
or.....
total number of ways=36
thank you very much........ i understood logic.......i think formula is
total number of gold coins!/at least 2 gold coins !*reaming coins!
correct me if i am wrong.....give me a correct formula.....
hi friend
i am not sure about general approach, i simply saw somewhere similar problem, and replicate the solution here
it may sound like combination with repetition:(n)C(k)=((k+n-1)!)/k!(n-1)!
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by aarati » Wed Aug 11, 2010 8:38 pm
clock60 wrote:
aarati wrote:
clock60 wrote:also got 36
at first give everybody 2 coins. and left with 13-2*6=7 coins
with no any restrictions 7 coins can be distributed among 3 persons
in 9!/(7!*2!)=36 ways
*******||.
or
***|*|***
or
*****||**
or.....
total number of ways=36
thank you very much........ i understood logic.......i think formula is
total number of gold coins!/at least 2 gold coins !*reaming coins!
correct me if i am wrong.....give me a correct formula.....
hi friend
i am not sure about general approach, i simply saw somewhere similar problem, and replicate the solution here
it may sound like combination with repetition:(n)C(k)=((k+n-1)!)/k!(n-1)!
thank you very much ...but how 7 gold coins can be distributed to 3 persons...can u explain me that once..
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