Question-32600

Question Number 32600 by Tinkutara last updated on 28/Mar/18

Commented by mrW2 last updated on 28/Mar/18

$${I}\:{think}\:{there}\:{are}\:\mathrm{15}!\:{ways}. \\ $$

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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