A salesman has five chocolates each of three different varieties. If

Question

A salesman has five chocolates each of three different varieties. If he has to sell 9 chocolates to 9 different people and he can sell at most two varieties of chocolates, in how many different ways can he sell the chocolates?

A. 1512
B. 756
C. 378
D. 252
E. 126

A. 1512
B. 756
C. 378
D. 252
E. 126

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nick1816 · Answer

Number of ways to choose 2 different varieties = 3C2

Now out of 9 persons, 5 have similar chocolates and other 4 have similar.

Number of ways can he sell the chocolates  = 3C2* \frac{9!}{4!5! }*2! = 3*9*7*2*2 = 756

abannore · Answer

Can you please explain why have you multiplied by 2!?

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Lipun · Answer

Hi  abannore ,

# of ways to choose 2 different chocolates out of 3 : 3C2
Each type has 5 chocolates. To obtain a sum of 9 using two types, we have to have 5 from 1 type and 4 from second type.
Number of possible arrangements for after choosing 2 types out of 3: 2!
(a) 5 from type A, 4 from type B
OR
(b) 5 from type B, 4 from type A
So, we need to multiply by 2!.

If you want to understand the other way around:
5A,4B
5A,4C
5B,4A
5B,4C
5C,4A
5C,4B
Total: 3! = 6

Hope it helps!

Thanks
Lipun

ScottTargetTestPrep · Answer

Solution:

The only way the salesman can sell at most two varieties of chocolate to 9 people is if he sells 5 chocolates of one kind and 4 chocolates of some other kind.

From the 9 people, we can choose 5 of them in 9C5 = 9!/(5!*4!) = (9 x 8 x 7 x 6)/(4 x 3 x 2) = 9 x 7 x 2 = 126 ways. The 5 people we chose will receive one variety of chocolate and the remaining 4 people will receive some other variety of chocolate.

There are 3 options for the variety of chocolate which will be sold to 5 people. There are 2 options for the variety of chocolate which will be sold to the remaining 4 people. Thus, we can choose the varieties of chocolate in 3 x 2 = 6 ways.

Combining the above findings, we conclude that the chocolates can be sold to the 9 people in 126 x 6 = 756 ways.

Answer: B

rajeshforyouu · Answer

3 Varieties - 5 Each

Selecting 2 out of 3 Varieties:  3C2

You need to pick 9 chocolates 
- 4 from Variety 1 & 5 from Variety 2 
or
- 4 from Variety 2 & 5 from Variety 1 =  2  ways

Selecting 4 people out of 9 -  9C4

Now you have  4 choc  of one Variety that will be distributed among 4 people &  5 choc  of another Variety that will be distributed among 5 people :  1  way

3C2*2*9C4*1 =   756

Option B

Spoorthii · Answer

I solved it this way but I'm not sure if this is correct. Please help!

Choose 2 varieties out of 3 = 3C2 = 3 ways
We now have 10 chocolates to be sold to 9 people which can be done in 10!/5! 5! = 252 ways

252*3 = 756

sanjitscorps18 · Answer

ScottTargetTestPrep ,  Bunuel

I needed your opinion, I started thinking on an extra layer of choice where I thought after choosing the 2 varieties, now I have 10 chocolates to choose from and 9 different people to distribute it to.

Do we also need to factor that I have to choose 9 chocolates out of 10 as an intermediary step. In the answers I see the next step to simply distribute (arrange) the chocolates to 9 different persons, but still had 10 to choose from?

Bunuel · Answer

In nick1816's solution that is accounted by 2! (5 of type X, 4 of type Y or 4 of type X, 5 of type Y).

A salesman has five chocolates each of three different varieties. If

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