mathmath333
  • mathmath333
Counting question
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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.
schrodinger
  • schrodinger
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Rushwr
  • Rushwr
??????? question ?
mathmath333
  • mathmath333
\(\large \color{black}{\begin{align} \normalsize \text{In how many ways can u distribute 7 identical gifts among 5 children.}\hspace{.33em}\\~\\ \end{align}}\)
mathmath333
  • mathmath333
ok

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More answers

ganeshie8
  • ganeshie8
say the stars are gifts : \[*~*~*~*~*~*~*\]
ganeshie8
  • ganeshie8
you want to split those 7 gifts into 5 parts, so place 4 bars in between them : \[*~*|~*|~*|~*~*|~*\]
ganeshie8
  • ganeshie8
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
ganeshie8
  • ganeshie8
see if you can tell what below arrangement represents : \[*|~*~*|~*|~*~*|~*\]
mathmath333
  • mathmath333
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
ganeshie8
  • ganeshie8
Perfect! notice, that string has 7 stars and 4 bars, so the total length of that string is 7+4 = 11
ganeshie8
  • ganeshie8
as you can see, the problem translates to finding the number of ways of choosing 4 positions for the bars from the 11 positions
mathmath333
  • mathmath333
I think now the answer is 11C4. as u are seleecting 11 objects in 4 ways
ganeshie8
  • ganeshie8
how many ways can you choose 4 different things(positions) from 11 different things(positions) ?
mathmath333
  • mathmath333
11C4
ganeshie8
  • ganeshie8
Thats it!
anonymous
  • anonymous
stars and bars indeed
ganeshie8
  • ganeshie8
Alternatively you could also think of it as forming different 11 letter words using 7 stars and 4 bars : 11!/(4!*7!)
mathmath333
  • mathmath333
\(\large \color{black}{\begin{align} & \normalsize \text{In how many ways can u distribute 7 identical gifts among 5 children.}\hspace{.33em}\\~\\ & \normalsize \text{such that each child gets at least 1 gift.}\hspace{.33em}\\~\\ \end{align}}\)
anonymous
  • anonymous
reserve 5 gifts, distribute the remaining 2, and then count the ways you can assign one of the reserved 5 to each child
ganeshie8
  • ganeshie8
you may use the same trick, consider 7 stars : \[*~*~*~*~*~*~*\]
ganeshie8
  • ganeshie8
you want to partition that into 5 nonempty parts, at what positions are you allowed to place the 4 bars ?
ganeshie8
  • ganeshie8
is below a valid arrangement ? \[*|~*|~*|~*~*|~*|~*\]
mathmath333
  • mathmath333
except the rear and front ends ?
mathmath333
  • mathmath333
yes,valid
ganeshie8
  • ganeshie8
im asking specifically if above arrangement is valid
ganeshie8
  • ganeshie8
good :)
ganeshie8
  • ganeshie8
how about below one : \[|*~*|~*~*~*|~*|~*\] what does it represent and is it a valid one ?
mathmath333
  • mathmath333
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
ganeshie8
  • ganeshie8
right, that means you don't like the first child what about below one : \[*~*|~|~*~*~*|~*|~*\]
mathmath333
  • mathmath333
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
ganeshie8
  • ganeshie8
so you cannot place bars next to each other and you cannot place bars on the ends
ganeshie8
  • ganeshie8
the only valid places for bars are : \[*~-*-~*-*-*-*-*\] those 6 dashes
ganeshie8
  • ganeshie8
four bars to place six positions to choose from how many total ways can u do it ?
mathmath333
  • mathmath333
6C4 ?
ganeshie8
  • ganeshie8
Yep!
ganeshie8
  • ganeshie8
a bit more generally, the number of positive integer solutions to the equation \(\large a+b+c+d+e=n\) is given by \(\large \dbinom{n-1}{4}\)
ganeshie8
  • ganeshie8
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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