Seven different coins are to be divided amongst three persons. If no

Question

Seven different coins are to be divided amongst three persons. If no two of the persons receive the same number of coins but each receives at least one coin & none is left over, then the number of ways in which the division may be made is

(A) 420
(B) 630
(C) 710
(D) 750
(E) 800

(A) 420 
(B) 630
(C) 710 
(D) 750
(E) 800

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ScottTargetTestPrep · Accepted Answer

Solution:
 
Let’s first assume that the coins are indistinguishable. Then the problem becomes: How many ways can 3 positive integers add up to 7 where no two integers are the same? We can see that we can only have:

1 + 2 + 4 = 7

Of course, here we are told that the coins are different or distinguishable. So let’s put that into consideration. Since the 7 coins are different, if the first person can only have 1 coin, he actually has 7 choices to pick the coin since they are all different. Then the next person has 6C2 ways to choose 2 coins from the remaining 6 if he is to pick 2 coins. Finally, the last person has 4C4 ways to choose his 4 coins from the remaining 4. Therefore, the total number of ways is:

7 x 6C2 x 4C4 = 7 x 15 x 1 = 105

However, the above only consider the case where the first person gets 1 coin, the second 2 coins and the third 4 coins. It may well be the case that the first person gets 2 coins, the second 4 coins and the third 1 coin. In that case, the total number of ways (using the same procedure as above) is:

7C2 x 5C4 x 1C1 = 21 x 5 x 1 = 105

As we can see, there are also 105 ways. Since there are 3! = 6 ways to rearrange the three numbers 1, 2, and 4, there are a total of 105 x 6 = 630 ways.
 
 Answer: B

pgwodehouse · Answer

Here is my solution:

Out of the 7 coins, one each is given to the three people. 
There are 7C3 ways of doing this.

Now, the remaining 4 coins can only be arranged in the combination of (0,1,3) to meet the condition that no two people have the same number of coins. 
# of ways of picking 3 coins = 4C3
After this there is only one way to pick the last coin

Since there are three people who can get these 3 coins, there are 3 ways to arrange this selection.

Hence the final answer is:
7C3(4C3*3*1) = 420 (A)

Gaurav2896 · Answer

Answer is b, 630

coins can be distributed as 
1,2,4 .
so no of ways = 3! *7c1*6c2*4c4 = 
630

GMATinsight · Answer

The ways to divide seven coins among 3 such that no tow receive the same number of coin can be {1,2, 4}

So we need to make selection of these three groups first which can be done in 7C4*3C2 = 35*3 = 105 ways

Then we need to arrange the three bunches among those three persons which can be done in = 3! = 6 ways

Total required outcomes = 105*6 = 630

Answer: Option B

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