Question 1195051
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Let's find out how many ways there are to pick the three men. 
n = 10 men
r = 3 selections
n C r = (n!)/(r!(n-r)!)
10 C 3 = (10!)/(3!*(10-3)!)
10 C 3 = (10!)/(3!*7!)
10 C 3 = (10*9*8*7!)/(3!*7!)
10 C 3 = (10*9*8)/(3!)
10 C 3 = (10*9*8)/(3*2*1)
10 C 3 = (720)/(6)
10 C 3 = 120


There are 120 ways to pick the three men from a pool of ten men. Order does not matter. This is why we use the nCr combination formula and not the nPr permutation formula.


Order doesn't matter on a committee like this because no member outranks another. No members hold special titles such as "president", "VP", "secretary", etc. Each seat is the same.


Follow similar steps for the women.
n = 10 women
r = 4 selections
n C r = (n!)/(r!(n-r)!)
10 C 4 = (10!)/(4!*(10-4)!)
10 C 4 = (10!)/(4!*6!)
10 C 4 = (10*9*8*7*6!)/(4!*6!)
10 C 4 = (10*9*8*7)/(4!)
10 C 4 = (10*9*8*7)/(4*3*2*1)
10 C 4 = (5040)/(24)
10 C 4 = 210


There are 210 ways to pick the four women from a pool of ten women. Like before, order doesn't matter.


Side note: The values 120 and 210 can be found in Pascal's Triangle. Look at the row that starts with 1,10,...
The 10C3 = 120 is the fourth item in that row, while 10C4 = 210 is the fifth item. The general item 10Cr is item number r+1.


Anyways, we found there are 120 ways to pick the men and 210 ways to pick the women.


Overall there are 120*210 = <u><font color=red>25200 ways</font></u> to form the seven person committee.
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