Question and Answers Forum

All Questions      Topic List

Permutation and Combination Questions

Previous in All Question      Next in All Question      

Previous in Permutation and Combination      Next in Permutation and Combination      

Question Number 134482 by benjo_mathlover last updated on 04/Mar/21

  How many ways can this be done if you distribute 25 identical pieces of candy among five children?

$$ \\ $$How many ways can this be done if you distribute 25 identical pieces of candy among five children?

Commented by mr W last updated on 04/Mar/21

you′ll get problem if you don′t give  each child 5 pieces! :)    mathematically to get an unique  answer you must specify if each  child may get nothing or must get  at least one piece. if i were you, i′ll  add if each child must get at least  two pieces.

$${you}'{ll}\:{get}\:{problem}\:{if}\:{you}\:{don}'{t}\:{give} \\ $$$$\left.{each}\:{child}\:\mathrm{5}\:{pieces}!\::\right) \\ $$$$ \\ $$$${mathematically}\:{to}\:{get}\:{an}\:{unique} \\ $$$${answer}\:{you}\:{must}\:{specify}\:{if}\:{each} \\ $$$${child}\:{may}\:{get}\:{nothing}\:{or}\:{must}\:{get} \\ $$$${at}\:{least}\:{one}\:{piece}.\:{if}\:{i}\:{were}\:{you},\:{i}'{ll} \\ $$$${add}\:{if}\:{each}\:{child}\:{must}\:{get}\:{at}\:{least} \\ $$$${two}\:{pieces}. \\ $$

Commented by benjo_mathlover last updated on 04/Mar/21

  hello sir, that's the problem written in the book  no other explanation

$$ \\ $$hello sir, that's the problem written in the book no other explanation

Commented by mr W last updated on 04/Mar/21

then it′s assumed that a box (here a  child) may be empty. in this case  there are C_4 ^(25+4) =((29!)/(25!4!))=23751 ways.

$${then}\:{it}'{s}\:{assumed}\:{that}\:{a}\:{box}\:\left({here}\:{a}\right. \\ $$$$\left.{child}\right)\:{may}\:{be}\:{empty}.\:{in}\:{this}\:{case} \\ $$$${there}\:{are}\:{C}_{\mathrm{4}} ^{\mathrm{25}+\mathrm{4}} =\frac{\mathrm{29}!}{\mathrm{25}!\mathrm{4}!}=\mathrm{23751}\:{ways}. \\ $$

Commented by mr W last updated on 04/Mar/21

if each child must get at least  two pieces, then we have in fact only  15 pieces to distribute, there are  C_4 ^(15+4) =3876 ways.

$${if}\:{each}\:{child}\:{must}\:{get}\:{at}\:{least} \\ $$$${two}\:{pieces},\:{then}\:{we}\:{have}\:{in}\:{fact}\:{only} \\ $$$$\mathrm{15}\:{pieces}\:{to}\:{distribute},\:{there}\:{are} \\ $$$${C}_{\mathrm{4}} ^{\mathrm{15}+\mathrm{4}} =\mathrm{3876}\:{ways}. \\ $$

Commented by benjo_mathlover last updated on 04/Mar/21

ok sir. thanks

$$\mathrm{ok}\:\mathrm{sir}.\:\mathrm{thanks} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com