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mathmath333

  • one year ago

Counting question

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  1. Rushwr
    • one year ago
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    ??????? question ?

  2. mathmath333
    • one year ago
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    \(\large \color{black}{\begin{align} \normalsize \text{In how many ways can u distribute 7 identical gifts among 5 children.}\hspace{.33em}\\~\\ \end{align}}\)

  3. mathmath333
    • one year ago
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    ok

  4. ganeshie8
    • one year ago
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    say the stars are gifts : \[*~*~*~*~*~*~*\]

  5. ganeshie8
    • one year ago
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    you want to split those 7 gifts into 5 parts, so place 4 bars in between them : \[*~*|~*|~*|~*~*|~*\]

  6. ganeshie8
    • one year ago
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    that arrangement represents : 2 gifts to first child 1 gift to second child 1 gift to third child 2 gifts to fourth child 1 gift to fifth child

  7. ganeshie8
    • one year ago
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    see if you can tell what below arrangement represents : \[*|~*~*|~*|~*~*|~*\]

  8. mathmath333
    • one year ago
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    1 gifts to first child 2 gift to second child 1 gift to third child 2 gifts to fourth child 1 gift to fifth child

  9. ganeshie8
    • one year ago
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    Perfect! notice, that string has 7 stars and 4 bars, so the total length of that string is 7+4 = 11

  10. ganeshie8
    • one year ago
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    as you can see, the problem translates to finding the number of ways of choosing 4 positions for the bars from the 11 positions

  11. mathmath333
    • one year ago
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    I think now the answer is 11C4. as u are seleecting 11 objects in 4 ways

  12. ganeshie8
    • one year ago
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    how many ways can you choose 4 different things(positions) from 11 different things(positions) ?

  13. mathmath333
    • one year ago
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    11C4

  14. ganeshie8
    • one year ago
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    Thats it!

  15. anonymous
    • one year ago
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    stars and bars indeed

  16. ganeshie8
    • one year ago
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    Alternatively you could also think of it as forming different 11 letter words using 7 stars and 4 bars : 11!/(4!*7!)

  17. mathmath333
    • one year ago
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    \(\large \color{black}{\begin{align} & \normalsize \text{In how many ways can u distribute 7 identical gifts among 5 children.}\hspace{.33em}\\~\\ & \normalsize \text{such that each child gets at least 1 gift.}\hspace{.33em}\\~\\ \end{align}}\)

  18. anonymous
    • one year ago
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    reserve 5 gifts, distribute the remaining 2, and then count the ways you can assign one of the reserved 5 to each child

  19. ganeshie8
    • one year ago
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    you may use the same trick, consider 7 stars : \[*~*~*~*~*~*~*\]

  20. ganeshie8
    • one year ago
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    you want to partition that into 5 nonempty parts, at what positions are you allowed to place the 4 bars ?

  21. ganeshie8
    • one year ago
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    is below a valid arrangement ? \[*|~*|~*|~*~*|~*|~*\]

  22. mathmath333
    • one year ago
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    except the rear and front ends ?

  23. mathmath333
    • one year ago
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    yes,valid

  24. ganeshie8
    • one year ago
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    im asking specifically if above arrangement is valid

  25. ganeshie8
    • one year ago
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    good :)

  26. ganeshie8
    • one year ago
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    how about below one : \[|*~*|~*~*~*|~*|~*\] what does it represent and is it a valid one ?

  27. mathmath333
    • one year ago
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    0 gifts to first child 2 gift to second child 3 gift to third child 1 gifts to fourth child 1 gift to fifth child invalid

  28. ganeshie8
    • one year ago
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    right, that means you don't like the first child what about below one : \[*~*|~|~*~*~*|~*|~*\]

  29. mathmath333
    • one year ago
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    2 gifts to first child 0 gift to second child 3 gift to third child 1 gifts to fourth child 1 gift to fifth child invalid

  30. ganeshie8
    • one year ago
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    so you cannot place bars next to each other and you cannot place bars on the ends

  31. ganeshie8
    • one year ago
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    the only valid places for bars are : \[*~-*-~*-*-*-*-*\] those 6 dashes

  32. ganeshie8
    • one year ago
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    four bars to place six positions to choose from how many total ways can u do it ?

  33. mathmath333
    • one year ago
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    6C4 ?

  34. ganeshie8
    • one year ago
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    Yep!

  35. ganeshie8
    • one year ago
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    a bit more generally, the number of positive integer solutions to the equation \(\large a+b+c+d+e=n\) is given by \(\large \dbinom{n-1}{4}\)

  36. ganeshie8
    • one year ago
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    similarly, the number of "non negative" integer solutions to the equation \(\large a+b+c+d+e=n\) is given by \(\large \dbinom{n+4}{4}\)

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