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?

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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?

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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any one you can help me
@welshfella can you slove it
its been a long time. I could do it by listing all the combinations I suppose but there must be a quicker way.
thanks for posting the question. As @welshfella says it gives us a chance to test our knowledge
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
There are a total of 18! (18 factorial possible permutations) here, but we shall work our way down!
The best way to think of this is to create 5 boxes! one for each marble
@BPDlkeme234 sorry to interrupt, there is a total of (5+5+3)=13 marbles only! :)
thatnkyou @mathlete, I had to make corrections anyway
so.... 13 factorial?
If the first box has to contain a blue marble then you have a choice of 5 marbles |dw:1439457071976:dw|
If the second box contains a green marble , you also have a choice of 5 marbles|dw:1439457195269:dw|
Now, since the third box must contain a black marble you have a choice of 3 marbles|dw:1439457353162:dw|
This leaves two choices and you can select any color for the last two choices. I will explain how many possibilities left
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|
Finally, having selected the 4th marble, you have 9 possible choices (13-4) |dw:1439457719650:dw|
This means the total number of ways he can arrange the case is: 5 x 5 x 3 x 10 x 9
So how did do @mathlete?
i think you are correct @BPDlkeme234 but not confirmed... Please verify another expert but thanks for help
@pooja195 can you help me
you could ask @satellite73 he is a moderator here
@satellite73 can you help me
i suck at these, but looks like the above answer is correct
\[5\times 5\times 3\times 10\times 9\] right?
6750 that is the answer you are sure
oh wait a second...
yeah that looks pretty good to me
@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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