DecentNabeel
  • DecentNabeel
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
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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DecentNabeel
  • DecentNabeel
@ganeshie8
DecentNabeel
  • DecentNabeel
@hartnn
DecentNabeel
  • DecentNabeel
@mathmate @mathmate

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More answers

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