Tom is arranging his marble collection in a collectors case. He has

Question

Tom is arranging his marble collection in a collectors case. He has five identical cat-eyes, five identical sulphides, and three identical agates. If he can fit exactly five marbles into the case and must have at least one of each type, how many different ways can he arrange the case?

(A) 120
(B) 150
(C) 420
(D) 1,260
(E) 1,680

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(A) 120
 (B) 150
 (C) 420
 (D) 1,260
 (E) 1,680

Bunuel · Accepted Answer

We have that 3 marbles  CSA  must be in the collection case as he must have at least one of each type.

Then five chosen marbles could be:

1.  CSA-XX  - meaning that other 2 marbles are the same (for example CSA- CC ):
 \frac{5!}{3!}*C^1_3=60  ( C^1_3  choosing which marble out of the 3 will be X and  \frac{5!}{3!}  arranging 5 marbles our of which 3 are identical).

2.  CSA-XY  - meaning that other 2 marbles are not the same (for example CSA- CS ):
 \frac{5!}{2!2!}*C^2_3=60  ( C^2_3  choosing which 2 marbles out of 3 will be X and Y and  \frac{5!}{2!2!}  arranging 5 marbles our of which 2 and other 2 are identical).

Total:  60+90=150 .

Answer: B.

shrouded1 · Answer

Given the constraints, there is 2 ways to choose the stones :

1) Take 2 each of 2 types and 1 of the third type
3 ways to choose the type for which 1 stone is to be taken
Ways to arrange = 5!/(2!x2!) = 30
Total = 30 x 3 = 90

2) Take 3 of 1 type, one each of the other types
3 ways to choose the type for which 3 stones are to be taken
Ways to arrange = 5!/3! = 20
Total = 20 x 3 = 60

So net number of ways = 150

Answer is (b)

chetan2u · Answer

hi..

so 5Cs, 5Ss and 3As.. 
atleast one of each C,S and A means TWO cases
 I.. 3 of one kind and one each of other two 
 \frac{5!}{3!}=20 
any one of 3 can be three, so 20*3=60
 II... 2 of two and one of third 
 \frac{5!}{2!2!}=30 
any one of 3 can be one, so 30*3=90

total 60+90=150

B

krikre · Answer

Could someone pls provide more comprehensive explanation?

lostin · Answer

Though explanation provided by Chetan is good, I am adding more details to his solution.

Tom has 5 cat-eys (say C) , 5 sulphides (say S) and 3 agates (say A). 
So Tom has 5C, 5S and 3A.

The case can fit 5 marbles and please note the question is asking for the arrangement so the order of marbles in Case is also important.
As per the question. We must have 1C, 1S and 1A.
so for remaining 2 spots -
We can Choose 2 of one kind or 1 of two kinds.

1. If we choose 2 of one kind -
We can choose 2 of one kind in 3 ways ( Either 2C or 2A or 2C)
So in total in case, we will have 3 of one kind and 2 other kinds of marble 
There is a formula to solve these kinds of arrangement but in case somebody doesn't recall -
We can arrange the 1 kind in 5 ways and second kind by 4. We have 3 identical marble left and 3 spots to fill so the order is not important now.
So ways = 5X4 = 20
so total ways as we can select first 2 marble in 3 ways 
= 5X4X3 = 60

2. If we choose 2 kinds of marble to fill 2 empty spaces -
Now we can choose marble in 3 ways ( 1A and 1C, 1A and 1S, 1S and 1C).
So in case, we will 2-2 marbles of 2 kinds and 1 marble of one kind.
Lets' assume we have 2A, 2S and 1C. 
for 1C, we can select spots in 5 ways.
for 2As, we can select spots in 6 ways (Select 2 out of 4, order not important).
for remaining 2S, we are left with 2 spots only after fixing 1C and 2As, so only one way.
So ways = 6X5 = 30 
so total ways as we can select first 2 marble in 3 ways 
= 30X3 = 90.

Total ways = Case 1 + Case 2 = 60 + 90 = 150

Note - It may be easier to solve with the formulas if you are aware of them.

Turkish · Answer

This has helped me a lot. Hopefully will help you. https://www.youtube.com/watch?v=QCq6qRJCTj8

krikre · Answer

Hello Lostin and Turkish, thank you for your responses.

Lostin could you kindly let me know what formulas you mean ?

Thank you.

lostin · Answer

Chetan has used the same formulas for solutions, very handy and fast in case of big numbers.

nirbhayg879 · Answer

ok so I will provide you answer in very simple way. Anyone can understand this.
let X represent cat eye, No. of X = 5
Y represent sulphides, No. of Y = 5
Z represent agates, No. of Z = 3
So here first we have to make combination of 5 out of total 13 and then we will arrange them.
these combinations are:-
1X 1Y 3Z 
1X 2Y 2Z
1X 3Y 1Z 
2X 1Y 2Z 
3X 1Y 1Z 
2X 2Y 1Z
Now we have to arrange(or make permutation of) each combination:-
Total No. of permutation we have:-
5!/3! + 5!/2!*2! + 5!/3! + 5!/2!*2! + 5!/3! + 5!/2!*2! = 150 = (B)

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