

there is a committee of 5 boys and 3 girls. at a meeting, 2 boys are absent. find the no. of ways they can sit at a round table with 8 seats.
ans is 2520
a deck of 10 cards is numbered from 0 to 9. find the no. of ways 5 cards can be chosen such that the sum of the chosen cards is larger than those not chosen.
ans is 126
how to dooooooooooo? pls give explanations! thx!



for the 1st question
the no of ways to arrange 6 people in 8 seats is 8P6
as they are seated in a circle we need to divide by 8
hence the ans 8P6 / 8 = 2520

for the 2nd question
there are 10C5 ways of choosing 5 cards without restriction
the 5 cards chosen is either bigger than or lesser than the cards not chosen
the probability that the chosen cards are greater than the "un chosen" cards is equal to the probability that the chosen cards are lesser than the "unchosen" cards
hence we can just take 10C5 divide by 2 = 126



Originally posted by quailmaster:
for the 1st question
the no of ways to arrange 6 people in 8 seats is 8P6
as they are seated in a circle we need to divide by 8
hence the ans 8P6 / 8 = 2520
Nice.
We can also factorial 8 different seats, but 2 are "repeated" as empty seats
Hence, the answer is
(8  1)! / 2! =2520
