Question Number 212522 by Nadirhashim last updated on 16/Oct/24
$$\:\:\boldsymbol{{in}}\:\boldsymbol{{how}}\:\boldsymbol{{many}}\:\boldsymbol{{ways}}\:\boldsymbol{{we}} \\ $$$$\boldsymbol{{can}}\:\boldsymbol{{distribute}}\:\mathrm{6}\:\boldsymbol{{distinct}} \\ $$$$\boldsymbol{{balls}}\:\boldsymbol{{in}}\:\mathrm{3}\:\boldsymbol{{identical}}\:\boldsymbol{{boxes}} \\ $$
Answered by mehdee7396 last updated on 16/Oct/24
$$\:\mathrm{6}/\mathrm{0}/\mathrm{0}\:{or}\:\mathrm{5}/\mathrm{1}/\mathrm{0}\:{or}\:\mathrm{4}/\mathrm{2}/\mathrm{0}\:\:{or}\:\mathrm{4}/\mathrm{1}/\mathrm{1}\:{or}\:\mathrm{3}/\mathrm{3}/\mathrm{0}/\:{or}\:\mathrm{3}/\mathrm{2}/\mathrm{1}\:{or}\:\mathrm{2}/\mathrm{2}/\mathrm{2} \\ $$$${C}\left(\mathrm{6},\mathrm{6}\right)+{C}\left(\mathrm{5},\mathrm{1}\right){C}\left(\mathrm{1},\mathrm{1}\right)+{C}\left(\mathrm{6},\mathrm{4}\right){C}\left(\mathrm{2},\mathrm{2}\right)+{C}\left(\mathrm{6},\mathrm{4}\right){C}\left(\mathrm{2},\mathrm{1}\right){C}\left(\mathrm{1},\mathrm{1}\right)+{C}\left(\mathrm{6},\mathrm{3}\right){C}\left(\mathrm{3},\mathrm{3}\right)+{C}\left(\mathrm{6},\mathrm{3}\right){C}\left(\mathrm{3},\mathrm{2}\right){C}\left(\mathrm{1},\mathrm{1}\right)+{C}\left(\mathrm{6},\mathrm{2}\right){C}\left(\mathrm{4},\mathrm{2}\right){C}\left(\mathrm{2},\mathrm{2}\right) \\ $$$$=\mathrm{275}\:\checkmark \\ $$
Commented by mr W last updated on 21/Oct/24
$${it}'{s}\:{wrong}\:{sir}. \\ $$$${the}\:{number}\:{of}\:{ways}\:{to}\:{distribute}\:{n} \\ $$$${distinct}\:{objects}\:{into} \\ $$$${n}_{\mathrm{1}} \:{groups}\:{with}\:{m}_{\mathrm{1}} \:{objects}\:{in}\:{each}\:{and} \\ $$$${n}_{\mathrm{2}} \:{groups}\:{with}\:{m}_{\mathrm{2}} \:{objects}\:{in}\:{each} \\ $$$${etc}. \\ $$$$\left({n}={n}_{\mathrm{1}} {m}_{\mathrm{1}} +{n}_{\mathrm{2}} {m}_{\mathrm{2}} +…+{n}_{{r}} {m}_{{r}} \right) \\ $$$${is}\:\frac{{n}!}{\underset{{k}=\mathrm{1}} {\overset{{r}} {\prod}}\left({m}_{{k}} !\right)^{{n}_{{k}} } {n}_{{k}} !} \\ $$$${therefore}\:{for}\:{example} \\ $$$$\mathrm{3}/\mathrm{2}/\mathrm{1}\:\Rightarrow\frac{\mathrm{6}!}{\mathrm{3}!\mathrm{2}!\mathrm{1}!}=\mathrm{60}={C}\left(\mathrm{6},\mathrm{3}\right){C}\left(\mathrm{3},\mathrm{2}\right){C}\left(\mathrm{1},\mathrm{1}\right) \\ $$$$\mathrm{2}/\mathrm{2}/\mathrm{2}\:\Rightarrow\frac{\mathrm{6}!}{\left(\mathrm{2}!\right)^{\mathrm{3}} \mathrm{3}!}=\mathrm{15}\neq{C}\left(\mathrm{6},\mathrm{2}\right){C}\left(\mathrm{4},\mathrm{2}\right){C}\left(\mathrm{2},\mathrm{2}\right) \\ $$
Answered by mr W last updated on 16/Oct/24
$${if}\:{the}\:{boxes}\:{may}\:{not}\:{be}\:{empty}: \\ $$$${S}\left(\mathrm{6},\:\mathrm{3}\right)=\mathrm{90}\:\checkmark \\ $$$${if}\:{the}\:{boxes}\:{may}\:{be}\:{empty}: \\ $$$${S}\left(\mathrm{6},\mathrm{3}\right)+{S}\left(\mathrm{6},\mathrm{2}\right)+{S}\left(\mathrm{6},\mathrm{1}\right)=\mathrm{90}+\mathrm{31}+\mathrm{1}=\mathrm{122}\:\checkmark \\ $$
Answered by mr W last updated on 17/Oct/24
$$\mathrm{6}/\mathrm{0}/\mathrm{0}\:\Rightarrow\mathrm{1}\:{way} \\ $$$$\mathrm{5}/\mathrm{1}/\mathrm{0}\:\Rightarrow{C}_{\mathrm{1}} ^{\mathrm{6}} =\mathrm{6}\:{ways} \\ $$$$\mathrm{4}/\mathrm{2}/\mathrm{0}\:\Rightarrow{C}_{\mathrm{2}} ^{\mathrm{6}} =\mathrm{15}\:{ways} \\ $$$$\mathrm{3}/\mathrm{3}/\mathrm{0}\:\Rightarrow{C}_{\mathrm{3}} ^{\mathrm{6}} /\mathrm{2}=\mathrm{10}\:{ways} \\ $$$$\mathrm{4}/\mathrm{1}/\mathrm{1}\:\Rightarrow\frac{\mathrm{6}!}{\mathrm{4}!\left(\mathrm{1}!\right)^{\mathrm{2}} \mathrm{2}!}=\mathrm{15}\:{ways} \\ $$$$\mathrm{3}/\mathrm{2}/\mathrm{1}\:\Rightarrow\frac{\mathrm{6}!}{\mathrm{3}!\mathrm{2}!\mathrm{1}!}=\mathrm{60}\:{ways} \\ $$$$\mathrm{2}/\mathrm{2}/\mathrm{2}\:\Rightarrow\frac{\mathrm{6}!}{\left(\mathrm{2}!\right)^{\mathrm{3}} \mathrm{3}!}=\mathrm{15}\:{ways} \\ $$$${totally}:\:\mathrm{122}\:{ways}\:\checkmark \\ $$