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Question Number 116883 by bemath last updated on 07/Oct/20
$$\mathrm{Find}\:\mathrm{the}\:\mathrm{number}\:\mathrm{m}\:\mathrm{of}\:\mathrm{ways}\:\mathrm{to} \\ $$$$\mathrm{partition}\:\mathrm{10}\:\mathrm{students}\:\mathrm{into}\:\mathrm{four}\:\mathrm{team} \\ $$$$\mathrm{so}\:\mathrm{that}\:\mathrm{two}\:\mathrm{team}\:\mathrm{contains}\:\mathrm{3}\:\mathrm{students} \\ $$$$\mathrm{and}\:\mathrm{two}\:\mathrm{team}\:\mathrm{contains}\:\mathrm{2}\:\mathrm{students}\:. \\ $$
Answered by john santu last updated on 07/Oct/20
$${There}\:{are}\:{m}'\:=\:\frac{\mathrm{10}!}{\mathrm{3}!.\mathrm{3}!.\mathrm{2}!.\mathrm{2}!}\:=\:\mathrm{25},\mathrm{200}\:{such}\:{ordered}\:{partition} \\ $$$${Since}\:{the}\:{teams}\:{form}\:{an}\:{unordered}\:{partition} \\ $$$${we}\:{divide}\:{m}'\:{by}\:\mathrm{2}!\:{because}\:{of}\:{the}\:{two} \\ $$$${cells}\:{with}\:\mathrm{3}\:{elements}\:{each}\:{and}\:\mathrm{2}!\: \\ $$$${because}\:{of}\:{the}\:{two}\:{cells}\:{with}\:\mathrm{2}\:{elements}\:{each} \\ $$$${Thus}\:{m}\:=\:\frac{\mathrm{25},\mathrm{200}}{\mathrm{2}!.\mathrm{2}!}\:=\:\mathrm{6300}\: \\ $$
Commented by bemath last updated on 07/Oct/20
$$\mathrm{yes}\:\mathrm{i}\:\mathrm{got}\:\mathrm{the}\:\mathrm{same}\:\mathrm{answer} \\ $$