Day 23: Last Minute Christmas Rush!

Probability Level 4

A disorganised mathematician is doing all his Christmas preparation in two days to make a party for his friends. He, being systematic, has made a list of seven activities: $A,B,C,D,E,F,$ and $G$ , to do (one at a time). However:

$A$ (buying Christmas food and presents) must be done before $B$ (preparing food).
$E$ (making math problems for his friends) must be done as one of the first four activities (because he thinks better in the morning).

For example, one valid way is $ABCEDFG$ . In total, how many ways are there for him to do the activities?

This problem is part of the set Advent Calendar 2014 .

The answer is 1440.

2 solutions

Michael Ng
Dec 22, 2014

We will consider condition 1 at the end. There are $4$ choices for E's position, then $6$ for for $A$ , $5$ for for $B$ then so on giving us $4 \times 6!=2880$ ways.

But then for every way where we have $A$ before $B$ , we have one other way with $A$ and $B$ switched around. So dividing by two gives the required answer: $\boxed{1440}$ .

How is there 4 choices for E, it has to be done in the morning so the position of E is fixed,that is at the top of the list...

aniket lal - 6 years, 5 months ago

$E$ must be done as one of the first four activities, so there are four choices for its position.

Michael Ng - 6 years, 5 months ago

ok...now I get it,thanks a lot.

aniket lal - 6 years, 5 months ago

Alex Pavinčič
Dec 26, 2014

There are 7! total arrangements of A-G with no conditions applied.

E must be in one of the first four spots, so only 4/7 * 7! potential arrangements remain.

A must be before B, which happens in half of the total arrangements. This event is independent of the previous event, so the number of arrangements remaining is 1/2 * 4/7 * 7! .

This is equivalent to 2 * 6! = 2 * 720 = 1440

Day 23: Last Minute Christmas Rush!

The answer is 1440.

2 solutions

0 pending reports