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

Commented by mrW2 last updated on 28/Mar/18

$I t h i n k t h e r e a r e 15! w a y s .$
Commented by Tinkutara last updated on 29/Mar/18

$B u t i f w e s e e l i k e t h i s^{15} C_{1} \times^{15} C_{1} +$ $^{14} C_{1} \times^{14} C_{1} +^{13} C_{1} \times^{13} C_{1} + . . . +^{1} C_{1} \times^{1} C_{1}$ $= 1^{2} + 2^{2} + 3^{2} + . . . + 15^{2} = 1240$ $I s i t w r o n g ?$
Commented by prakash jain last updated on 29/Mar/18

$M . M . M . M . M . M . M . M . M . M . M . M . M . M . M .$ $There are 15 . s . You can 15 women$ $in 15! ways to create 15 teams .$ $Square Method 3 M (abc), 3 W (def)$ $^{3} C_{1} \times^{3} C_{1} = 9$ $ad bd cd$ $ae be ce$ $af bf cf$ $The above 9 do not comprise$ $of all teams (1 M, 1 W) so do not consitute$ $a valid solution .$
Commented by Tinkutara last updated on 29/Mar/18

$B u t i f t h e r e w e r e o n l y 3 m e n a n d 3$ $w o m e n a n d w e h a v e t o m a k e 3 t e a m s$ $e a c h h a v i n g 1 m e n a n d 1 w o m e n,$ $T o t a l w a y s = 1^{2} + 2^{2} + 3^{2} = 14$
Commented by Joel578 last updated on 29/Mar/18

$3 men : A, B, C$ $3 woman : a, b, c$ $A a, B b, C c \to 1$ $A a, B c, C b \to 2$ $A b, B a, C c \to 3$ $A b, B c, C a \to 4$ $A c, B b, C a \to 5$ $A c, B a, C b \to 6$ $There are 3! = 6 total ways$
Commented by Tinkutara last updated on 29/Mar/18
Thank you very much Sir! I got the answer. ��
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