A box contains six tiles: A, B, C, D, E and F.?

Question

How many possible 6-letter permutations can you draw from the box?

Answer 1

sevenhundredandtwenty

Answer 2

When you pick letters from the box, the first pick involves 6 letters.
The second pick involves 5 letters, third involves 4 letters, until there is one letter left and one possibility.

Multiply the number of possibilities from each draw together and you have:

6*5*4*3*2*1 = 720 permutations. (its written as 6!, meaning 6-factorial)

Answer 3

30

Answer 4

Without Replacement: 6! = 6*5*4*3*2*1

With Replacement = 6^6 = 6*6*6*6*6*6

Answer 5

raj, your clarification replaced into short, neat and easy yet you purely made one little mistake that you need to favor to edit the cost for the 4,4 determination you wrote as 2 * (6C4) even if it will be (6C4)^2 giving zerolucat's answerw of 465. Joe, your approach feels like it really is sensible yet you're double counting. frequently those 2 balls you p.c.. from the superb 6 are an similar you've picked interior the unique 6. it somewhat is why your decision is a lot larger than the actual answer

Answer 6

The answer is 1 (assuming you don't put any of the tiles back after drawing them).

That aside, I guess you could do an infinite # of sketches.

Answer 7

I think it's <6 to the power of 6>.
6x6x6x6x6x6
No that's wrong!

Answer 8

6! permutations.
6x5x4x3x2x1=720 permutations.

Answer 9

Thinking creatively.


You could turn the E on its side and make either an M or a W.

So 3

Answer 10

6! = six factorial = 720

Answer 11

Only one, then there aren't any tiles left in the box.

DOH!