Timothy is re-arranging his marble collection.he has five identical blue marbles,five identical green marbles and three identical black marbles he can fit exactly five marbles into a case and must have at least one of each.How many different ways can he arrange the case in?

- DecentNabeel

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- DecentNabeel

@ganeshie8

- DecentNabeel

@hartnn

- DecentNabeel

@mathmate @mathmate

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- DecentNabeel

any one you can help me

- DecentNabeel

@welshfella can you slove it

- welshfella

its been a long time. I could do it by listing all the combinations I suppose but there must be a quicker way.

- anonymous

thanks for posting the question. As @welshfella says it gives us a chance to test our knowledge

- anonymous

I am going to summarize the problem as follows:
5 identical blue marbles
5 identical green marbles
3 identical black marbles
he can fit 5 identical marbles into a case and must have at least one of each
How many different ways can he arrange the case in

- anonymous

There are a total of 18! (18 factorial possible permutations) here, but we shall work our way down!

- anonymous

The best way to think of this is to create 5 boxes! one for each marble

- mathmate

@BPDlkeme234
sorry to interrupt, there is a total of (5+5+3)=13 marbles only! :)

- anonymous

thatnkyou @mathlete, I had to make corrections anyway

- UsukiDoll

so.... 13 factorial?

- anonymous

If the first box has to contain a blue marble then you have a choice of 5 marbles
|dw:1439457071976:dw|

- anonymous

If the second box contains a green marble , you also have a choice of 5 marbles|dw:1439457195269:dw|

- anonymous

Now, since the third box must contain a black marble you have a choice of 3 marbles|dw:1439457353162:dw|

- anonymous

This leaves two choices and you can select any color for the last two choices. I will explain how many possibilities left

- anonymous

you have selected 3 marbles already leaving a total of , thanks to @mathlete informing me, 10 marbles. So you then have 10 possible choices you can make!|dw:1439457604678:dw|

- anonymous

Finally, having selected the 4th marble, you have 9 possible choices (13-4) |dw:1439457719650:dw|

- anonymous

This means the total number of ways he can arrange the case is:
5 x 5 x 3 x 10 x 9

- anonymous

So how did do @mathlete?

- DecentNabeel

i think you are correct @BPDlkeme234
but not confirmed...
Please verify another expert
but thanks for help

- DecentNabeel

@pooja195 can you help me

- anonymous

you could ask @satellite73 he is a moderator here

- DecentNabeel

@satellite73 can you help me

- anonymous

i suck at these, but looks like the above answer is correct

- anonymous

\[5\times 5\times 3\times 10\times 9\] right?

- DecentNabeel

6750 that is the answer you are sure

- anonymous

oh wait a second...

- anonymous

yeah that looks pretty good to me

- mathmate

@decentnabeel
Since you seem to be still at it, here's my take:
We need to place 5 marbles in the case, with at least one colour of each, there can only be a maximum of 3 marbles of a single colour. Thus with 5 blue, 5 green and 3 black, we have enough for any combination, and we need not worry about the 13 marbles any more.
Let B=blue, G=green, K=black.
Looking at the shelf, the colour combinations are either
case 1: (3,1,1) of each colour, or
There are three ways to get these colour combinations, i.e.
(3B,G,K), (3G,B,K) and (3K,B,G).
For each colour combination, the number of arrangements (permutations) is 5!/(3!1!1!)
for a total of 3(5!/(3!1!1!)) arrangements
case 2: (2,2,1) of each colour.
There are three ways to get these colour combinations, i.e.
(2B,2G,K), (2G,2K,B), (2K,2B,G).
For each colour combination, the number of arrangements (permutations) is 5!/(2!2!1!)
for a total of 3(5!/(2!2!1!)) arrangements
So the total number of arrangements is the sum of cases 1 & 2, namely
3(5!/(3!1!1!))+3(5!/(2!2!1!))

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