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DecentNabeel

  • one year ago

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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  1. DecentNabeel
    • one year ago
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    @ganeshie8

  2. DecentNabeel
    • one year ago
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    @hartnn

  3. DecentNabeel
    • one year ago
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    @mathmate @mathmate

  4. DecentNabeel
    • one year ago
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    any one you can help me

  5. DecentNabeel
    • one year ago
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    @welshfella can you slove it

  6. welshfella
    • one year ago
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    its been a long time. I could do it by listing all the combinations I suppose but there must be a quicker way.

  7. anonymous
    • one year ago
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    thanks for posting the question. As @welshfella says it gives us a chance to test our knowledge

  8. anonymous
    • one year ago
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    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

  9. anonymous
    • one year ago
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    There are a total of 18! (18 factorial possible permutations) here, but we shall work our way down!

  10. anonymous
    • one year ago
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    The best way to think of this is to create 5 boxes! one for each marble

  11. mathmate
    • one year ago
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    @BPDlkeme234 sorry to interrupt, there is a total of (5+5+3)=13 marbles only! :)

  12. anonymous
    • one year ago
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    thatnkyou @mathlete, I had to make corrections anyway

  13. UsukiDoll
    • one year ago
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    so.... 13 factorial?

  14. anonymous
    • one year ago
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    If the first box has to contain a blue marble then you have a choice of 5 marbles |dw:1439457071976:dw|

  15. anonymous
    • one year ago
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    If the second box contains a green marble , you also have a choice of 5 marbles|dw:1439457195269:dw|

  16. anonymous
    • one year ago
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    Now, since the third box must contain a black marble you have a choice of 3 marbles|dw:1439457353162:dw|

  17. anonymous
    • one year ago
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    This leaves two choices and you can select any color for the last two choices. I will explain how many possibilities left

  18. anonymous
    • one year ago
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    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|

  19. anonymous
    • one year ago
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    Finally, having selected the 4th marble, you have 9 possible choices (13-4) |dw:1439457719650:dw|

  20. anonymous
    • one year ago
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    This means the total number of ways he can arrange the case is: 5 x 5 x 3 x 10 x 9

  21. anonymous
    • one year ago
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    So how did do @mathlete?

  22. DecentNabeel
    • one year ago
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    i think you are correct @BPDlkeme234 but not confirmed... Please verify another expert but thanks for help

  23. DecentNabeel
    • one year ago
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    @pooja195 can you help me

  24. anonymous
    • one year ago
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    you could ask @satellite73 he is a moderator here

  25. DecentNabeel
    • one year ago
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    @satellite73 can you help me

  26. anonymous
    • one year ago
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    i suck at these, but looks like the above answer is correct

  27. anonymous
    • one year ago
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    \[5\times 5\times 3\times 10\times 9\] right?

  28. DecentNabeel
    • one year ago
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    6750 that is the answer you are sure

  29. anonymous
    • one year ago
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    oh wait a second...

  30. anonymous
    • one year ago
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    yeah that looks pretty good to me

  31. mathmate
    • one year ago
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    @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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