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# Counting

last edited by 13 years, 2 months ago

# Problem 1 of 3

Jessica, Richard and 6 of their friends attend a movie.

(a) In how may ways can they be seated in a row so that Jessica and Richard do not sit next to each other?

(b) In how many of these ways are Jessica and Richard sitting at either end of the row?

(c) After the movie they all went out to eat. They were seated at a round table. Again, Jessica and Richard will not sit next to each other. In how many ways can everyone be seated at the table?

Solution

Sorry i totally forgot to take the "still working on this" thing off... well i hope everything below is good... if not, well i did my best and i guess that's all i can do.. good luck to everyone, and "Don't we just love this wiki math stuff??" :D

a)

There are 8 kids in total going to the movie. Therefore there are 8! ways to arrange all the kids (without restrictions).

8! = 40320 Total ways everyone can be arranged (no restrictions)

If Jessica and Richard were going to sit together, we would throw them into a bag. There would now be only 7 things to arrange (6 kids, and 1 bag 2 kids inside). That means there are 7! was to arrange the 6 kids and bag. Now we must open the bag and let Jessica and Richard out, now there are 2! ways to arrange them. 7! ways to do the first thing and 2! ways to do the second so we will multiply them.

7! x 2! =10080 Which of course is all the ways that Jessica and Richard can sit together.

Now i realize the question says "not sit together", but now that i've found out how many ways they can sit together this problem is much easier. Now all that we have to do is take the total number of ways students can sit (without restrictions) and subtract the ways jessica and Richard can sit together.

8! - (7! x 2!) = 30240 Which is the number of ways that the kids can be arranged where Jessica and Richard do not sit together.

b)

This question is just different. In order to make this question easy to understand, we can use a visual representation.

_ _ _ _ _ _ _ _

2 6 5 4 3 2 1 1

The first slot can only be filled in one of two ways (Jessica or Richard). The next slot can be filled by any one of the 6 students that are not Jessica or Richard. The next slots (slots from the left 3, 4, 5, 6, and 7) can then be filled with the remaining 5 students who are not Jessica or Richard. Finally the last slot can be filled with the one of the two (Jessica or Richard) that did not

sit at the left end of the row.

So....

2 x 6 x 5 x 4 x 3 x 2 x 1 x 1 = 1440

which is the number of ways that Jessica and Richard can sit on either end of the row of students.

C)

This question is very similar to the question above (a) where it is easier to answer the compliment question then use it to find the answer to the given question. So here we go!!

There are a total of 8 kids going to eat after the movie. These 8 kids will be seated at a round table. 8 students means 8! ways to sit right?? Well normally yes, but because they are being seated at a ROUND table, the first person that sits is a reference point for the rest of the students. This counting means that the reference point (first person to sit) does not count. So subtract 1 student from the total.

8 - 1 = 7 students

There are 7 students that must be arranged now, meaning that the students can be arranged 7! ways, without restrictions.

7! = 5040 all ways for the 8 students to sit around the table (no restrictions)

Now we will first figure out all the ways that Richard and Jessica can sit together. To do this we must place them in a bag. If we do this there are then 7 items to arrange (6 kids and the bag 2 kids inside). This means there are 7! ways right?? Nope again, we have to remember to subtract one from the original for the reference point. Therefore there are 6! ways to arrange the students and the bag. Then we must open the bag, and there are 2! ways to seat Jessica and Richard together. 6! ways to do the first thing and 2! ways to do the second. So, that’s right, we mulitiply.

6! x 2! = 1440 All the possible ways to arrange the students around the table where Jessica and Richard DO sit together.

Now we have to find out the answer to the original question by subtracting the ways Jessica and Richard CAN sit together from all of the possible ways the students can sit around that table.

7! - (6!x 2!) = 3600 All the possible ways the students can sit around the table where Jessica and Richard DO NOT sit together.

# Problem 2 of 3

The 6th term of the binomial expansion of (x3 - 4/x)n contains x16.

(a) Find the value of n.

(b) How many terms are there in the full expansion?

(c) Does the expansion have a constant term? If so, which one is it?

Solution

# Problem 3 of 3

Solve for n in each of the following:

(a) nC2 = 10

(b) nC3 = 3(nP2)

(c) nP3 = 2(nC4)

Solution