Combination HELP PLEASE!!?

Question

Four people are to be selected at random from a group of four couples. In how many ways can this be done, given the following conditions?
a. there are no restriction.. i got this to be.. 8 C 4 = 70
b. the group must have at least one couple..
c. each couple must be represented in the group.

i don't know how to do b and c... can anyone help

Answer 1

b. Number of ways of selecting a couple = 4 C 1 =4
Number of ways of selecting another two from the remaining 6 = 6 C 2 = 15
=>Number of combinations = 4 x 15 = 60

c. Number of ways of selecting a certain person from a couple = 2C1 = 2
Number of couples = 4
=>Number of combinations = 2 ^ 4 =16
(Just like tossing a coin four times)

Answer 2

b. since the group must have at least one couple, then after you choose one couple, the rest of the possible selections would be 6 C 2 = 15. But since there are four couples, so there are 4 ways to choose the couple, the total would be 4 * 15 = 60.

c. If each couple has to be represented, then one person must be chosen from each couple. Since there are two different ways to choose a person from each of the four couples, the total choices would be 2*2*2*2 = 16

Answer 3

a. There is no restriction.
(8C4) = 70

You are correct.

c. Each couple must be represented in the group.
(2C1)^4 = 2^4 = 16

b. The group must have at least one couple.

If you think about it, that means the group will not have a couple. It's all the combinations not in c.

70 - 16 = 54

The answer is NOT 60. The posters above did not account for the fact that some of the combinations have two couples and so are double counted.

Answer 4

a) this is right 70 = 8X7X6X5divided by (4X3X2X1)

b) one couple In order to have at least one couple, you need to figure out the possibilites of no couples and subtract it from part a: To do this the choices are 8 X 6 X 4 X2 divided by (4X3X2X1) the first pick you have 8 choices: once you choose a person you need to delete his/her spouse, so the second pick leaves 6 choices, the third pick 4 choices and the last pick 2 choices.
that answer is 16 (8X6X4X2 diveded by 4 X 3 X 2 X1)
for no couples. 70 (no restrictions - part a) - 16 (no couples) = 54 (at least one couple)

c) each couple must be in the group: you have 2 choices times 2 choices times 2 choices times 2 choices this equals 16.

It also would be the same as no couples: 8 choices times 6 choices times 4 choices times 2 choices (each time you pick a person you must eliminate his/her spouse so you have room for one of each couple. , and then divided by 4 X 3 X2 X1. This would also be 16 combinations

Answer 5

b. case of exactly 1 couple: there are 4 ways to select a couple, remaining 2 people must be from different couples, number of ways to select them is:
6C2 - number of ways where the 2 are a couple
= 15 - number of remaining couples
= 15 - 3
= 12,
subtotal: (4)(12) = 48,
case of 2 couples: 4C2 = 6, subtotal: 6,
total: 48 + 6 = 54;

c. all 4 couples must be used & each has 2 possibilities, so answer is 2^4 = 16.

Answer 6

8x2=16 16x4=?

Answer 7

AC-DC=BB KEEP TRYING THE TRICK IS
KNOWING YOUR (A) FROM (B)