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.
how to solve this type of questions..............
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4GMAT_Mumbai
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clock60
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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
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.
thank you very much........ i understood logic.......i think formula isclock60 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
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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shekhar.kataria
- Master | Next Rank: 500 Posts
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How can 7 coins can be distributed among 3 persons in 9!/(7!*2!) ways, Please explain, confused hereclock60 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
any other way approaching this question?
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
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
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
-
clock60
- Legendary Member
- Posts: 759
- Joined: Mon Apr 26, 2010 10:15 am
- Thanked: 85 times
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hi friendaarati wrote:thank you very much........ i understood logic.......i think formula isclock60 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
total number of gold coins!/at least 2 gold coins !*reaming coins!
correct me if i am wrong.....give me a correct formula.....
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..clock60 wrote:hi friendaarati wrote:thank you very much........ i understood logic.......i think formula isclock60 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
total number of gold coins!/at least 2 gold coins !*reaming coins!
correct me if i am wrong.....give me a correct formula.....
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)!
