i. Mahowny rolls a 6-sided die five times, and the product of his rolls is 1200. How many different sequences of rolls could there have been? (The order of the rolls matters.)

ii. There are 18 chairs numbered from 1 through 18 around a circular table. How many ways can three people be seated, so that no two people are adjacent?

iii. Jeff the fly starts at the point in the coordinate plane. At each step, Jeff takes a step to the right, left, up, or down. After steps, how many different points could Jeff end up at?

I appreciate your help! Thank you!

Guest Feb 11, 2020

#1**+1 **

Here's my attempt for the first one :

i. Mahowny rolls a 6-sided die five times, and the product of his rolls is 1200. How many different sequences of rolls could there have been? (The order of the rolls matters.)

1200 factors as 2^4 * 3 * 5^2 = 2 * 2 * 2 * 2 * 3 * 5 * 5 = 4 * 4 * 3 * 5 * 5

So....the number of different *identifiable* roll sequenes =

5! 120

_______ = ____ = 30

2! * 2! 4

CPhill Feb 11, 2020

#3**0 **

This is the answer they gave, which I got 2nd try:

The prime factorization of 1200 is 2^4 * 3 * 5^2. The only way to get a factor of 5 is a roll of 5 itself, so two of the rolls must have been a 5. That means the product of the other three rolls were

The only ways to get a factor of 3 is through the rolls of 3 or 6. If one of the three remaining rolls is a 3, then the product of the other two rolls is 16. The only way this can happen is if the other two rolls were 4 and 4. If one of the three remaining rolls is a 6, then the product of the other two rolls is 8. The only way this can happen is if the other two rolls were 2 and 4.

So, the rolls must have been either 5, 5, 3, 4, 4 in some order, or 5, 5, 6, 2, 4 in some order. For 5, 5, 3, 4, 4, there are 5!/(2!*2!) =30

possible sequences. For 5, 5, 6, 2, 4, there are 5!/2! =60

possible sequences. This gives us a total of 90 possible sequences.

Guest Feb 12, 2020