You are going to make a password using 6 letters and 5 numbers without repeating anything. How many passwords can you make that begin with the letter E​? D, S, E, X, Q ,I ,K and 5 6 2 3 1 9

Answer 1

This is a combinatorics problem. The first thing to decide is whether we are dealing with a permutation or combination. Because the order of the characters matters ("EDSXQI562319" is certainly a different password than "EIQXSD913265," for example) we have a permutation.

First, let's consider the letters. The first letter must be "E," so we don't need to worry about it. You can think of that position as having 1 possibility.

Now we need five more letters, but we have six (D, S, X, Q, I, K) from which to choose. Therefore, we want to do "6 pick 5": 6*5*4*3*2 = 720.

Next we need 5 numbers, and there are 5 (5, 6, 2, 3, 1, 9) from which to choose. Therefore, we want to do "5 pick 5": 5*4*3*2*1 = 120.

Finally, the fundamental counting principle says that we multiply these results together, so the final answer is 1*720*120 = 86,400.