SOLUTION: A group of 30 people consists of 15 women and 15 men.
How many ways are there to
(a) arrange the entire group in a row? (2)
(b) form a row of 6 people from the men? (2)
(c) for
Algebra ->
Permutations
-> SOLUTION: A group of 30 people consists of 15 women and 15 men.
How many ways are there to
(a) arrange the entire group in a row? (2)
(b) form a row of 6 people from the men? (2)
(c) for
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Question 1154161: A group of 30 people consists of 15 women and 15 men.
How many ways are there to
(a) arrange the entire group in a row? (2)
(b) form a row of 6 people from the men? (2)
(c) form the group into a row with all the women in front of all the men? (3)
(d) choose a man and a women from the group? (3)
(e) choose four women and five men from the group? (3)
(f) choose 10 people from the group? (3)
(g) pair the women and men off? (3)
(h) form 10 pairs from the group? (3)
(i) divide the group into two groups (group 1 and group 2) of equal size? (3)
(j) divide the group into 2 equal groups, where each group in its own has as many men as
women in it? (3)
(k) divide the groups into two groups of equal size such that group 1 contains at least 4
men? (3)
(l) divide the group into two groups, each having size at least one? (3)
Each of the 30 choose one cooldrink from 10 types:
(m) How many ways could they do this? (3)
(n) How many different combinations of cooldrink could be ordered from the cafeteria,
where you don’t care who gets which but only want to get the number of each type
correct? (3)
Given that the group orders
15 of type 1
6 of type 2
3 of type 3
6 of type 4.
(o) In how many different ways can I distribute the cooldrinks amongst the 30 people, if I
do not care who ordered what? (4)
(p) What is the probability that a random division of the cooldrinks amongst the people
give each what they ordered? Answer by ikleyn(53937) (Show Source):
