Combinatorics, probability and statistics

Re: Combinatorics, probability and statistics

The first student can be in anyone of 35 seats. The second in any one of 34 seats, etc. So, you have this many permutations: 35 * 34 * ...* 7 * 6 = 35!/5!, or more generally n!/(n-k)! where n is the number of seats and k is the number of students. The exclamation point, of course, represents the factorial of the number.

Each handshake is a result of two out of sixteen people meeting. Order doesn't matter in this case (it doesn't matter if it's Mary and Joe vs Joe and Mary) so you are looking for the number of unordered pairs where n=16 (combinations, or k = 2) The general formula for combinations is n!/(k!(n-k)!).

For example, if there were 6 people at the party (n=6) the number of unordered pairs (k=2) is 6!/(2!(6-2)!) or 6!/(2!*4!) = 6*5*4*3*2*1 = 720/(2*24) = 720/48 = 15.

I'll let you do the math for n=16 (let me know what you get).



A normal distribution has the same mean, median, and mode. Given that the distribution is normally distributed, you know that the mean is the center of the interval between the min and the max (256-244 = 12/2 =6 or 250). All of the values encompass the area from the mean +/- four SD, or 250 +/- 6. An easy estimate of the sd is 6/4 = 1.5

Thanks a lot guys