Question 765042: A certain class has 8 boys and 10 girls. How many different sets of four class officers:president, vice president, treasurer and secretary, cant be formed from students in this class if the president and treasurer must be girls and the vice president and secretary must be boys? Please explain!
Answer by ramkikk66(644)   (Show Source):
You can put this solution on YOUR website!
President and treasurer must be girls. So you need to choose 2 girls from 10,
which can be done in C(10,2) ways = (10*9)/(1*2) = 45 ways.
For each of these combinations, there are 2 ways in which the 2 girls can
be allotted the 2 posts (e.g. Girl1 can be the Pres and Girl2 the treasurer,
and vice versa).
Hence possible combinations = 45*2
Similarly, VP and secretary must be boys. So you need to choose 2 boys from 8,
which can be done in C(8,2) ways = (8*7)/(1*2) = 28 ways.
For each of these combinations, there are 2 ways in which the 2 boys can
be allotted the 2 posts (e.g. Boy1 can be the VP and Boy2 the secretary,
and vice versa).
Hence possible combinations = 28*2
Finally, each set of 2 girls can be paired with each set of 2 boys to form the
final set of officers. So the number of possible combinations = 45*2*28*2 =
 ways.
Hope you got it :)
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