Permutations and Combinations...?

Question

There are ten desks in a row. (a) In how many ways can 5 students be seated at consecutive desks? (Hint: First determine the number of ways that the 5 consecutive desks be chosen). (b) If the five students are seated for a test and there is to be exactly one empty desk between each two students, in how many ways can the students be seated in a row?

Answer 1

(a) Different ways 5 students can be seated at 5 consecutive desks is found by a simple permutation: 5! = 120. Since there are 10 desks in a row, this permutation can start at desk 1 (seated 1~5) all the way up to desk 6 (6~10). So you multiply 120 by 6 = 720.

(b) Likewise, you start with the simple permutation and get 120. With this configuration, you can have students seated on odd numbered seats (1,3,5,7,9) or even numbered seats (2,4,6,8,10). So you multiply 120 by 2 = 240.

EDIT: Thanks sbjnyc (above) for first having the wrong answer then copying my answer. *hmpfff* EDIT #2: No worries sbjnyc. While you're at it, you may want to edit your answer again. The two patterns are OXOXOXOXOX and XOXOXOXOXO (as of 7/20, you have both patterns looking exactly the same).

Answer 2

a) If there are 10 desks in a row then there are 6 ways of selecting 5 consecutive desks (desks 1-5, desks 2-6, desks 3-7, desks 4-8, desks 5-9 and desks 6-10).

There are 5! ways of arranging 5 students in 5 desks (1 of 5 in the first 1 of 4 in the second, 1 of 3 in the 3rd 1 if 2 in the 4th and 1 in the 5th).

So the # of ways of arranging 5 students is 6*5! = 6! = 720

b) the basic patterns are OXOXOXOXOX or OXOXOXOXOX where O is a filled desk and X is an empty desk. There are no other arrangements possible.

There are 5! ways of arranging 5 students in each pattern.

So the total number of ways of arranging the students is

2 * 5! = 240

Edit: I didn't copy you namja but it may seem that way as I usually go through my answers a couple of times to see if I made any errors and this time I caught 2. I take my time while editing so if a bunch of answers come up during that time it can look a bit odd.

I'd rather do it this way though then just typing some placeholder for position as some do. It would be nice if YA sorted the answers by "last edited" time to make things more honest.

Answer 3

enable her choose for the ribbons one via one. For the 1st one, she has 6 strategies. For the 2d, there are 5 strategies left. So there are 30 diverse techniques of choosing an ordered pair of ribbons. inspite of the incontrovertible fact that, Cathy does not care correct to the order. Taking first a blue after which a white ribbon is the comparable as taking a white ribbon first after which a blue one. this skill we counted double the style of opportunities. So the actual variety is 5*6 / 2 = 15. it is, in result, a mixture: 2C6. The reasoning in the back of the formulation for combinations is as follows: you may placed all six ribbons in 6! diverse orderings. Of the six appeared after ribbons you will then pick the 1st 2. As you do no longer care correct to the order of those 2, you may divide the variety via the style of techniques of ordering those 2: 2!. the comparable is going for the 4 ribbons Cathy did no longer pick: those could be in 4! diverse orderings. So the full style of achieveable combinations is 6! / (2! 4!)

Answer 4

This sounds like a question from someone's take-home test or homework assignment.

Have you looked for the answer in the textbook?