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Question 1) 5 friends go to a movie, and sit in the same row side by side. How many different ways can this be done if Bob and Cindy sit next to each other?

My answer: 72

I got this answer by doing 4!*2= 48 and then 5!-48 = 72

After looking back at my notes I think I made a mistake and did an extra step in my work. Meaning that the answer should just be 48.

Question 2) 10 students are assigned to grade an exam for their class. The coordinator wants to randomly assign them into 2 different teams. One of 7 and one of 3. How many different ways can this be done?

My answer: 14,400

I had no idea how to do this one. But I started with (10 choose 7)*(10 choose 3) which got me  14,400.

I'm thinking this would be a combination problem rather than a permutation problem because it stated that the students are randomly selected.

My next idea for this problem is that it would be 7!/3!= 840

This is mainly just guesswork though.
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This is really combinatorics, not statistics.

>I got this answer by doing 4!*2= 48 and then 5!-48 = 72

>After looking back at my notes I think I made a mistake and did an extra step in my work. Meaning that the answer should just be 48.

I'm not sure why you did 5!-48 either, but 4!·2! is correct (treat Bob and Cindy as one element).

>Question 2) 10 students are assigned to grade an exam for their class. The coordinator wants to randomly assign them into 2 different teams. One of 7 and one of 3. How many different ways can this be done?

Hint: really all you need to count is how many ways to form the group of 7 (or how many ways to form the group of 3, which is equivalent). Because then you have no choice for how to form the other group. If this is still too hard, start with smaller numbers, e.g. 5 people that you have to split in to a group of 3 and a group of 2, to grasp the logic. Then you can check your answer by listing and counting all possibilities.
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It's 4×2!=8 for Bob and Cindy and

3!=6 for the others.

8×6=48 is the answer
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