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Question-32600




Question Number 32600 by Tinkutara last updated on 28/Mar/18
Commented by mrW2 last updated on 28/Mar/18
I think there are 15! ways.
$${I}\:{think}\:{there}\:{are}\:\mathrm{15}!\:{ways}. \\ $$
Commented by Tinkutara last updated on 29/Mar/18
But if we see like this^(15) C_1 ×^(15) C_1 +  ^(14) C_1 ×^(14) C_1 +^(13) C_1 ×^(13) C_1 +...+^1 C_1 ×^1 C_1   =1^2 +2^2 +3^2 +...+15^2 =1240  Is it wrong?
$${But}\:{if}\:{we}\:{see}\:{like}\:{this}\:^{\mathrm{15}} {C}_{\mathrm{1}} ×^{\mathrm{15}} {C}_{\mathrm{1}} + \\ $$$$\:^{\mathrm{14}} {C}_{\mathrm{1}} ×^{\mathrm{14}} {C}_{\mathrm{1}} +^{\mathrm{13}} {C}_{\mathrm{1}} ×^{\mathrm{13}} {C}_{\mathrm{1}} +…+^{\mathrm{1}} {C}_{\mathrm{1}} ×^{\mathrm{1}} {C}_{\mathrm{1}} \\ $$$$=\mathrm{1}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} +\mathrm{3}^{\mathrm{2}} +…+\mathrm{15}^{\mathrm{2}} =\mathrm{1240} \\ $$$${Is}\:{it}\:{wrong}? \\ $$
Commented by prakash jain last updated on 29/Mar/18
M.M.M.M.M.M.M.M.M.M.M.M.M.M.M.  There are 15 .s. You can 15 women  in 15! ways to create 15 teams.  Square Method 3 M (abc), 3W(def)  ^3 C_1 ×^3 C_1 =9  ad   bd   cd  ae    be   ce  af     bf    cf  The above 9 do not comprise  of all teams (1M,1W) so do not consitute  a valid solution.
$$\mathrm{M}.\mathrm{M}.\mathrm{M}.\mathrm{M}.\mathrm{M}.\mathrm{M}.\mathrm{M}.\mathrm{M}.\mathrm{M}.\mathrm{M}.\mathrm{M}.\mathrm{M}.\mathrm{M}.\mathrm{M}.\mathrm{M}. \\ $$$$\mathrm{There}\:\mathrm{are}\:\mathrm{15}\:.\mathrm{s}.\:\mathrm{You}\:\mathrm{can}\:\mathrm{15}\:\mathrm{women} \\ $$$$\mathrm{in}\:\mathrm{15}!\:\mathrm{ways}\:\mathrm{to}\:\mathrm{create}\:\mathrm{15}\:\mathrm{teams}. \\ $$$$\mathrm{Square}\:\mathrm{Method}\:\mathrm{3}\:\mathrm{M}\:\left(\mathrm{abc}\right),\:\mathrm{3W}\left(\mathrm{def}\right) \\ $$$$\:^{\mathrm{3}} \mathrm{C}_{\mathrm{1}} ×\:^{\mathrm{3}} \mathrm{C}_{\mathrm{1}} =\mathrm{9} \\ $$$$\mathrm{ad}\:\:\:\mathrm{bd}\:\:\:\mathrm{cd} \\ $$$$\mathrm{ae}\:\:\:\:\mathrm{be}\:\:\:\mathrm{ce} \\ $$$$\mathrm{af}\:\:\:\:\:\mathrm{bf}\:\:\:\:\mathrm{cf} \\ $$$$\mathrm{The}\:\mathrm{above}\:\mathrm{9}\:\mathrm{do}\:\mathrm{not}\:\mathrm{comprise} \\ $$$$\mathrm{of}\:\mathrm{all}\:\mathrm{teams}\:\left(\mathrm{1M},\mathrm{1W}\right)\:\mathrm{so}\:\mathrm{do}\:\mathrm{not}\:\mathrm{consitute} \\ $$$$\mathrm{a}\:\mathrm{valid}\:\mathrm{solution}. \\ $$
Commented by Tinkutara last updated on 29/Mar/18
But if there were only 3 men and 3  women and we have to make 3 teams  each having 1 men and 1 women,  Total ways=1^2 +2^2 +3^2 =14
$${But}\:{if}\:{there}\:{were}\:{only}\:\mathrm{3}\:{men}\:{and}\:\mathrm{3} \\ $$$${women}\:{and}\:{we}\:{have}\:{to}\:{make}\:\mathrm{3}\:{teams} \\ $$$${each}\:{having}\:\mathrm{1}\:{men}\:{and}\:\mathrm{1}\:{women}, \\ $$$${Total}\:{ways}=\mathrm{1}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} +\mathrm{3}^{\mathrm{2}} =\mathrm{14} \\ $$
Commented by Joel578 last updated on 29/Mar/18
3 men: A, B, C  3 woman: a, b, c  Aa, Bb, Cc  → 1  Aa, Bc, Cb  → 2  Ab, Ba, Cc  → 3  Ab, Bc, Ca  → 4  Ac, Bb, Ca  → 5  Ac, Ba, Cb  → 6  There are 3! = 6 total ways
$$\mathrm{3}\:\mathrm{men}:\:{A},\:{B},\:{C} \\ $$$$\mathrm{3}\:\mathrm{woman}:\:{a},\:{b},\:{c} \\ $$$${Aa},\:{Bb},\:{Cc}\:\:\rightarrow\:\mathrm{1} \\ $$$${Aa},\:{Bc},\:{Cb}\:\:\rightarrow\:\mathrm{2} \\ $$$${Ab},\:{Ba},\:{Cc}\:\:\rightarrow\:\mathrm{3} \\ $$$${Ab},\:{Bc},\:{Ca}\:\:\rightarrow\:\mathrm{4} \\ $$$${Ac},\:{Bb},\:{Ca}\:\:\rightarrow\:\mathrm{5} \\ $$$${Ac},\:{Ba},\:{Cb}\:\:\rightarrow\:\mathrm{6} \\ $$$$\mathrm{There}\:\mathrm{are}\:\mathrm{3}!\:=\:\mathrm{6}\:\mathrm{total}\:\mathrm{ways} \\ $$
Commented by Tinkutara last updated on 29/Mar/18
Thank you very much Sir! I got the answer. ��������

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