[janemorales]

Did I figure this correctly?

A puzzle in the newspaper presents a matching problem. The names of 10 in one U.S. presidents are listed column, and their vice presidents are listed in random order in the second column. The puzzle asks the reader to match each president with his vice president. If you make the matches randomly, how many matches are possible? What is the probability all 10 of your matches are correct?
10Pres x 10 Vpres = 100 possible matches
10/100 = 10%

[galactus]

It's a lot more than 100, Jane. Think about it. You have 10 choices for the first, 9 for the second, 8 for the third, and so on. That is 10! Ways.

[Unknown008]

Yes, that's correct. If you match the first president with the first vpresident, you don't have 9 but 9! (ie 9x8x7x6x5x4x3x2x1) ways. In total, you'll have 10! (ie 10x9x8x7x6x5x4x3x2x1). The probability now, you can do it? And note that there is only one correct answer for each president. Therefore, there is only one combination among the so many random, wrong combinations.

[janemorales]

So if I do 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 = 362,880

this is correct?

[ebaines]

So if I do 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 = 362,880

this is correct?

Not quite - look at Galactus's response again. The first Pres has 10 possible VPs he could be combined with; after that the 2nd Pres has 9 possible VPs left, etc. So the total number of combinations is 10! = 10 x 9 x 8 x... x 2 x 1

[Unknown008]

I said IF you chose the first president, there is 9! Number of possibilities. In all, THERE ARE 10! Combinations because you may start with any president.