Probability question help (Check my answers please) Q. selection ordered pairs nPr, nCr

Question

Question : A television director has to schedule commercials during 6 time slots allocated to commercials during the national telecast of the first period of a Canadiens hockey game.

(a) In how many ordered ways can the television director schedule 6 different commercials during the 6 time slots?

(b) In how many ways can the television director fill the 6 time slots allocated to commercials, if there are 4 different commercials, of which a given one is to be shown 3 times while each of the others is to be shown once?

(c) As in (b) except the commercial shown three times can’t be shown before or after itself (i.e., not consecutively). How many ways can the television director fill the 6 time slots?

My answers:

a) 6! ways (6P6)

b) 4 commercials, 1 is to be repeated 3 times.

R _ _ _ + _ R _ _ + _ _ R _ + _ _ _ R
… R stands for this slot contains 3 repeated adverts

The other 3 slots can be chosen in 3P3 ways = 3!

Thus, it is 1 x 3! + 1 x 3! + 1 x 3! + 1 x 3! = 24 ways

c) c) Repeated advertisements cannot be show before or after itself

This can be arranged in two ways.

R _ R _ R _                                Or                        _ R _ R _ R

Which is 

1 x 3! + 1 x 3! = 12 ways.

Answer 1

In how many ways can six different commercials be shown in six time slots?

Your answer $6!$ is correct.

In how many ways can four different commercials, one of which is to be shown three times, be shown in six time slots?

Strategy:

1. Choose three of the six time slots for the repeated commercial.
2. Arrange the other three commercials in the remaining three time slots?

There are $\binom{6}{3}$ ways to select three of the six time slots for the repeated commercial and $3!$ to arrange the remaining three distinct commercials in the remaining three time slots, so there are $$\binom{6}{3}3! = \frac{6!}{3!3!} \cdot 3! = \frac{6!}{3!}$$ The factor of $3!$ in the denominator represents the number of ways the commercial that is repeated could be permuted within an arrangement without producing an arrangement that is distinguishable from the given arrangement.

What did you do wrong?

By treating the repeated commercial as a block, you only counted those cases in which the repeated commercial ran in three consecutive slots. However, there is no such restriction.

In how many ways can four different commercials, one of which is to be shown three times, be shown in six time slots if the repeated commercial cannot be shown in consecutive slots.

We will arrange three blue balls, one green ball, one red ball, and one yellow ball so that no two of the blue balls are consecutive. The blue balls represent the positions of the repeated commercial, no two of which are consecutive. The other balls represent the positions of the other three commercials.

Strategy:

1. Arrange the green, red, and yellow balls in a row.
2. This creates four spaces, two between successive balls and two at the ends of the row in which a blue ball can be placed.
3. In order to separate the blue balls, we must choose three of these four spaces in which to insert a blue ball.

The green, red, and yellow balls can be arranged in a row in $3!$ ways. This creates four spaces in which we can place the blue balls. For instance, we could have the arrangement $$\square g \square r \square y \square$$ To ensure that no two blue balls are consecutive, we must choose three of these four spaces in which to place a blue ball, which can be done in $\binom{4}{3}$ ways. Hence, the number of admissible schedules is $$3!\binom{4}{3}$$

What did you do wrong?

You assumed that the repeated commercials must alternate with the other commercials. However, this is not the case since the arrangements $bgrbyb$ and $bgbryb$ are both admissible since no two of the blue balls are consecutive.

Answer 2

For b) $AAABCD$ represents the $4$ commercials with A shown $3$ times. The number of permutations is $\frac{6!}{3!} = 120$.

For c) $AxAxAx, AxxAxA, AxAxxA, xAxAxA$ are the only $4$ possible configurations for $A$ whereby $3!$ are the permutations for $BCD$.

Hence $4\cdot 3! = 24$.