Probability help?

Question

how many different ways can five people A,B,C,D,E, sit in a row if

a) C must sit to the right of, but not necessarily next to B
B) D and E must not sit together

pls include solutions and explanations

Answer 1

a) Without the restriction there are 5! = 120 ways to sit people in a row. 1/2 of these will have C to the left of B and 1/2 will have C to the right of B. So the answer is 60 ways.

b) Without the restriction there are 5! = 120 ways to sit people in a row. There are 4 pairs of seats next to each other. D and E could sit in each of these pairs in 2 ways for a total of 8. For every 8 cases there are 3! ways to seat the other people for a grand total of 6*8 = 48 seatings that we must avoid with the net possible solutions honoring this restriction being 120-48 = 72 ways.

Answer 2

A)

BCxxx = 3! = 6 permutations
BxCxx = 3! = 6 permutations
BxxCx = 3! = 6 permutations
BxxxC = 3! = 6 permutations
xBCxx = 3! = 6 permutations
xBxCx = 3! = 6 permutations
xBxxC = 3! = 6 permutations
xxBCx = 3! = 6 permutations
xxBxC = 3! = 6 permutations
xxxBC = 3! = 6 permutations

total = 60 permutations


B)

DxExx
DxxEx
DxxxE
xDxEx
xDxxE
xxDxE

three are 36 permutations of this form and 36 with E before D for a total of 72 permutations.

Answer 3

5

Answer 4

There are 5! ways to arrange the seating without constraints, right? Well, for each arrangement there is another with B and C reversed and the other seats unchanged. This means that exactly half of the 5! arrangements will pass condition (a).

One way to get to condition (b) is to note that 2/5 of the arrangements will have D seated at one end or the other, and of these, only 3 out of 4 seatings will have E at least 2 seats away. Independently, 3/5 of the arrangements will have D in one of the middle 3 seats, and for each of these only 2 of the 4 remaining seats will be the required distance away.

(a) and (b) are independent conditions, so you can multiply the *fractions* of the population they represent (this is really just the product rule for the "and" of two independent probabilities) and then multiply by the population size (5!) to get the final answer.

Answer 5

If they can sit in any order, then there is 120 ways=5!
For a)
If B sits right next to C, or BC
then there are 24 ways=4! that they can sit like that.
If BOC where O is someone else,
18 ways
BOOC=12
BOOOC=6
Add all together, get 60. Probability is 1/2
B) When DE=24 ways,
ED=24 ways.
Add is 48 ways.
120-48=72
72/120=3/5
Probability is 3/5.

Answer 6

qa
since lap-sitting isn't allowed, B can only either be right or left of C. half the time exactly, so 0.5 is there for you.

0.5 * 5P5
= 0.5 * 5!
= 60

qb
DXEEE = 3 possible seats for E.
XDXEE = 2
XXDXE = 1
so 3+2+1 there.
but those only for D being E's left.
EXDDD
XEXDD
XXEXD
so i double them. (times 2)
obviously 3P3 is to arrange ABC.

2 * 3P3 * (3+2+1)
= 2*3!*6
= 72