What is the minimum value for n for which (1*2*3...*n) will be 990.
Also pl find the attachment
5-Day Free Trial
5-day free, full-access trial TTP
Available with Beat the GMAT members only code
MORE DETAILS
pl solve ths..exam next week
This topic has expert replies
-
gmat20092009
- Junior | Next Rank: 30 Posts
- Posts: 28
- Joined: Sun Jul 19, 2009 6:41 am
-
prindaroy
- Master | Next Rank: 500 Posts
- Posts: 113
- Joined: Thu Jul 16, 2009 11:23 am
- Thanked: 15 times
- GMAT Score:730
For the answer to your second question (the one with the attachment);
It's 16P2 - 16 = 16!/14! - 16 = 240 - 16 = 224
Now 16P2 is essentially the total number of ways in which mike and the other guy can be arranged. 16 ways is the number of ways in which they ALWAYS sit across each other. There are 5 + 3 possibilities where Mike takes his seat first and Jordan sits across. There are then 5 + 3 possibilities when Jordan sits first and Mike sits across. Hence 5+3+5+3 = 16.
Now for your first question, I don't think that's the entire question. It's probably what is the minimum value of n for which n! = 1*2*....*n is a multiple of 990. In which case the answer is 11.
It's 16P2 - 16 = 16!/14! - 16 = 240 - 16 = 224
Now 16P2 is essentially the total number of ways in which mike and the other guy can be arranged. 16 ways is the number of ways in which they ALWAYS sit across each other. There are 5 + 3 possibilities where Mike takes his seat first and Jordan sits across. There are then 5 + 3 possibilities when Jordan sits first and Mike sits across. Hence 5+3+5+3 = 16.
Now for your first question, I don't think that's the entire question. It's probably what is the minimum value of n for which n! = 1*2*....*n is a multiple of 990. In which case the answer is 11.
