Subscribe Feed (RSS)Who can solve these problems about counting rules?
July 18th, 2009 by admin | Filed under Mathematics.
john p asked:
1.In how many different orders can six people sit around a round table.
2.For a group of 4 men and 5 women, how many committees of size 3 are possible: a) with no restriction? b) With 1 man and 2 women?
3.How many six-person volleyball teams be made from a group of seven men and three women if:
a.There is no sex discrimination?
b.The team must contain at least two women?
4.How many bridge hands are possible containing 4 spades, 6 diamonds, 1 club and 2 hearts?
5.A president and a treasurer are to be chosen from a student club consisting of 50 people. Ho many different choices of officers are possible if
a.there are no restrictions
b.A will serve only if he is president
c.B and C will serve together or not at all?
6.Find the number of ways in which nine contestants can be ranked first, second, third, fourth and fifth to their heights.
7.How many different permutations are there of the letters from the word Mississippi?
1.In how many different orders can six people sit around a round table.
2.For a group of 4 men and 5 women, how many committees of size 3 are possible: a) with no restriction? b) With 1 man and 2 women?
3.How many six-person volleyball teams be made from a group of seven men and three women if:
a.There is no sex discrimination?
b.The team must contain at least two women?
4.How many bridge hands are possible containing 4 spades, 6 diamonds, 1 club and 2 hearts?
5.A president and a treasurer are to be chosen from a student club consisting of 50 people. Ho many different choices of officers are possible if
a.there are no restrictions
b.A will serve only if he is president
c.B and C will serve together or not at all?
6.Find the number of ways in which nine contestants can be ranked first, second, third, fourth and fifth to their heights.
7.How many different permutations are there of the letters from the word Mississippi?
Tags: Bridge Hands, Seven Men, Three Women
Yep, I can too.
1) for circle arrangement, n object can be arrange using (n-1)!..(u need to fix 1 person and arrange others)
2)a)9!
b) 4C1*5C2
3)a)10C6
b) sum of team with 2 women, and 3 women. 7C4*2C2 +
7C3*3C3
4) 12C4*12C6*12C1*12C2
5) a)50C2
b) dunno
c) dunno
6) 2 ways, 1st, the heighst is on the right side, 2nd, the heights on the left side
7)n!/a!b!….. where a!b! the number of same object
11!/4!4!2!
1) It is just like putting 6 people into 6 desks, except the first person only has one choice. The first choice doesn’t matter because the table is round. So, instead of 6!, the answer is 5!
2)
a) No restriction: 9 choose 3 or 9*8*7/(3*2*1)
b) Pick 1 man out of 4 and 2 women out of 5: 4C1 * 5C2
3)
a) Just like 2a. 10C6
b) Just like 2b except, add in the teams with 3 women.. 3C2 women * 7C4 men + 3C3 women * 7C3 men.
4) 13 cards in a bridge hand. 13 cards in each suit.
13C4 spades * 13C6 diamonds * 13C1 club *13C2 hearts
13!/(4! * 9!) * 13!/(6! * 7!) * 13!/(1!* 13!) * 13!/(2! * 11!)
5)
a) Order matters so the answer isn’t just 50C2. Instead, you have 50 choices for the president and only 49 for the treasurer, 50*49.
b) Ignore A and you get 49*48. Now, assume A is picked (1 choice) and you have 49 choices for treasurer. 49*48 + 1*49.
c) Ignore B and C and you get 48*47. Now assume that B and C got their wish, there are 2*1 ways they can be happy. 48*47 + 2*1.
6) I’m not sure what difference the height is supposed to make if you assume that they each have a different fixed height. Since each group of 5 only has 1 way in which they can be arranged by height, the answer would be 9C5. However, if you read it as their heights can be treated as a variable, you have 9 choices for the first, 8 choices for the second,… or 9*8*7*6*5 different arrangements of 5 people out of a group of nine. This is the same as how many different ways can 9 people be seated in 5 desks sitting in a row.
7) There are 11 letters in Mississippi. The initial response is 11!. BUT WAIT, there are 4 ’s’, 4 ‘i’, and 2 ‘p’ as repeats. The order of these letters creates duplicates. So, divide by 4!, 4!, and 2!. 11!/(4! * 4! * 2!).