0

For the question below, I have worked out the solution for a) in which I've got:

$9\cdot8\cdot7\cdot6\cdot5 = 15120$.

However, I am struggling to solve part b) of the question.

Any help would be appreciated!

The question:

Nine boxes are each labelled with a different natural number from 1 to 9. Five people are allowed to take one box each. In how many different ways can this be done if:

a) There are no restrictions 
b) The first three people decide that they will take even numbered boxes?
dave
  • 705
  • 4
  • 16
FriendlyMan18
  • 3

1 Answers1

1

The first one has 4 options (on of all even numbers, i.e. $\left\{ 2,4,6,8 \right\} $), the second one has 3 options (one of all even numbers left), the third one has 2 options, the forth one has 6 options (one of all numbers left), the fifth one has 5 options.

So we get $4\cdot3\cdot2\cdot6\cdot5 = 6!$

dave
  • 705
  • 4
  • 16