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Question Number 212522 by Nadirhashim last updated on 16/Oct/24
in how many ways we can distribute 6 distinct balls in 3 identical boxes
$$\:\:\boldsymbol{{in}}\:\boldsymbol{{how}}\:\boldsymbol{{many}}\:\boldsymbol{{ways}}\:\boldsymbol{{we}} \\ $$$$\boldsymbol{{can}}\:\boldsymbol{{distribute}}\:\mathrm{6}\:\boldsymbol{{distinct}} \\ $$$$\boldsymbol{{balls}}\:\boldsymbol{{in}}\:\mathrm{3}\:\boldsymbol{{identical}}\:\boldsymbol{{boxes}} \\ $$
Answered by mehdee7396 last updated on 16/Oct/24
6/0/0 or 5/1/0 or 4/2/0 or 4/1/1 or 3/3/0/ or 3/2/1 or 2/2/2 C(6,6)+C(5,1)C(1,1)+C(6,4)C(2,2)+C(6,4)C(2,1)C(1,1)+C(6,3)C(3,3)+C(6,3)C(3,2)C(1,1)+C(6,2)C(4,2)C(2,2) =275 ✓
$$\:\mathrm{6}/\mathrm{0}/\mathrm{0}\:{or}\:\mathrm{5}/\mathrm{1}/\mathrm{0}\:{or}\:\mathrm{4}/\mathrm{2}/\mathrm{0}\:\:{or}\:\mathrm{4}/\mathrm{1}/\mathrm{1}\:{or}\:\mathrm{3}/\mathrm{3}/\mathrm{0}/\:{or}\:\mathrm{3}/\mathrm{2}/\mathrm{1}\:{or}\:\mathrm{2}/\mathrm{2}/\mathrm{2} \\ $$$${C}\left(\mathrm{6},\mathrm{6}\right)+{C}\left(\mathrm{5},\mathrm{1}\right){C}\left(\mathrm{1},\mathrm{1}\right)+{C}\left(\mathrm{6},\mathrm{4}\right){C}\left(\mathrm{2},\mathrm{2}\right)+{C}\left(\mathrm{6},\mathrm{4}\right){C}\left(\mathrm{2},\mathrm{1}\right){C}\left(\mathrm{1},\mathrm{1}\right)+{C}\left(\mathrm{6},\mathrm{3}\right){C}\left(\mathrm{3},\mathrm{3}\right)+{C}\left(\mathrm{6},\mathrm{3}\right){C}\left(\mathrm{3},\mathrm{2}\right){C}\left(\mathrm{1},\mathrm{1}\right)+{C}\left(\mathrm{6},\mathrm{2}\right){C}\left(\mathrm{4},\mathrm{2}\right){C}\left(\mathrm{2},\mathrm{2}\right) \\ $$$$=\mathrm{275}\:\checkmark \\ $$
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_1 groups with m_1 objects in each and n_2 groups with m_2 objects in each etc. (n=n_1 m_1 +n_2 m_2 +...+n_r m_r ) is ((n!)/(Π_(k=1) ^r (m_k !)^n_k n_k !)) therefore for example 3/2/1 ⇒((6!)/(3!2!1!))=60=C(6,3)C(3,2)C(1,1) 2/2/2 ⇒((6!)/((2!)^3 3!))=15≠C(6,2)C(4,2)C(2,2)
$${it}'{s}\:{wrong}\:{sir}. \\ $$$${the}\:{number}\:{of}\:{ways}\:{to}\:{distribute}\:{n} \\ $$$${distinct}\:{objects}\:{into} \\ $$$${n}_{\mathrm{1}} \:{groups}\:{with}\:{m}_{\mathrm{1}} \:{objects}\:{in}\:{each}\:{and} \\ $$$${n}_{\mathrm{2}} \:{groups}\:{with}\:{m}_{\mathrm{2}} \:{objects}\:{in}\:{each} \\ $$$${etc}. \\ $$$$\left({n}={n}_{\mathrm{1}} {m}_{\mathrm{1}} +{n}_{\mathrm{2}} {m}_{\mathrm{2}} +…+{n}_{{r}} {m}_{{r}} \right) \\ $$$${is}\:\frac{{n}!}{\underset{{k}=\mathrm{1}} {\overset{{r}} {\prod}}\left({m}_{{k}} !\right)^{{n}_{{k}} } {n}_{{k}} !} \\ $$$${therefore}\:{for}\:{example} \\ $$$$\mathrm{3}/\mathrm{2}/\mathrm{1}\:\Rightarrow\frac{\mathrm{6}!}{\mathrm{3}!\mathrm{2}!\mathrm{1}!}=\mathrm{60}={C}\left(\mathrm{6},\mathrm{3}\right){C}\left(\mathrm{3},\mathrm{2}\right){C}\left(\mathrm{1},\mathrm{1}\right) \\ $$$$\mathrm{2}/\mathrm{2}/\mathrm{2}\:\Rightarrow\frac{\mathrm{6}!}{\left(\mathrm{2}!\right)^{\mathrm{3}} \mathrm{3}!}=\mathrm{15}\neq{C}\left(\mathrm{6},\mathrm{2}\right){C}\left(\mathrm{4},\mathrm{2}\right){C}\left(\mathrm{2},\mathrm{2}\right) \\ $$
Answered by mr W last updated on 16/Oct/24
if the boxes may not be empty: S(6, 3)=90 ✓ if the boxes may be empty: S(6,3)+S(6,2)+S(6,1)=90+31+1=122 ✓
$${if}\:{the}\:{boxes}\:{may}\:{not}\:{be}\:{empty}: \\ $$$${S}\left(\mathrm{6},\:\mathrm{3}\right)=\mathrm{90}\:\checkmark \\ $$$${if}\:{the}\:{boxes}\:{may}\:{be}\:{empty}: \\ $$$${S}\left(\mathrm{6},\mathrm{3}\right)+{S}\left(\mathrm{6},\mathrm{2}\right)+{S}\left(\mathrm{6},\mathrm{1}\right)=\mathrm{90}+\mathrm{31}+\mathrm{1}=\mathrm{122}\:\checkmark \\ $$
Answered by mr W last updated on 17/Oct/24
6/0/0 ⇒1 way 5/1/0 ⇒C_1 ^6 =6 ways 4/2/0 ⇒C_2 ^6 =15 ways 3/3/0 ⇒C_3 ^6 /2=10 ways 4/1/1 ⇒((6!)/(4!(1!)^2 2!))=15 ways 3/2/1 ⇒((6!)/(3!2!1!))=60 ways 2/2/2 ⇒((6!)/((2!)^3 3!))=15 ways totally: 122 ways ✓
$$\mathrm{6}/\mathrm{0}/\mathrm{0}\:\Rightarrow\mathrm{1}\:{way} \\ $$$$\mathrm{5}/\mathrm{1}/\mathrm{0}\:\Rightarrow{C}_{\mathrm{1}} ^{\mathrm{6}} =\mathrm{6}\:{ways} \\ $$$$\mathrm{4}/\mathrm{2}/\mathrm{0}\:\Rightarrow{C}_{\mathrm{2}} ^{\mathrm{6}} =\mathrm{15}\:{ways} \\ $$$$\mathrm{3}/\mathrm{3}/\mathrm{0}\:\Rightarrow{C}_{\mathrm{3}} ^{\mathrm{6}} /\mathrm{2}=\mathrm{10}\:{ways} \\ $$$$\mathrm{4}/\mathrm{1}/\mathrm{1}\:\Rightarrow\frac{\mathrm{6}!}{\mathrm{4}!\left(\mathrm{1}!\right)^{\mathrm{2}} \mathrm{2}!}=\mathrm{15}\:{ways} \\ $$$$\mathrm{3}/\mathrm{2}/\mathrm{1}\:\Rightarrow\frac{\mathrm{6}!}{\mathrm{3}!\mathrm{2}!\mathrm{1}!}=\mathrm{60}\:{ways} \\ $$$$\mathrm{2}/\mathrm{2}/\mathrm{2}\:\Rightarrow\frac{\mathrm{6}!}{\left(\mathrm{2}!\right)^{\mathrm{3}} \mathrm{3}!}=\mathrm{15}\:{ways} \\ $$$${totally}:\:\mathrm{122}\:{ways}\:\checkmark \\ $$

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