# Probability

• Jun 17th 2013, 12:36 PM
alexander9408
Probability
1. How many even numbers can be formed if the digits of the number 112433 are rearranged?
2. In arranging a 10-day examination time-table involving 10 subjects and one subject per day, a teacher plans to have Mathematics, Physics and Chemistry all separate by at least one day. How many ways are possible?

Anyone please explain these to me, thank you.
PS: The answer for question 1 is 120 and the answer for question 2 is 1693440.
• Jun 17th 2013, 12:58 PM
Plato
Re: Probability
Quote:

Originally Posted by alexander9408
1. How many even numbers can be formed if the digits of the number 112433 are rearranged?

The answer for question 1 is 120

To be even the string must end in 2 or 4.

The string $\displaystyle 11433$ can be arranged in $\displaystyle \frac{5!}{(2!)^2}=30$ ways.
Add a 2 to the right-hand end of each of those 30 strings.
That is is an even number.

Now swap the 4 and the 2 from each of above that gives another 30.

So 60 in all.

BTW: there are 120 odd numbers.
• Jun 17th 2013, 12:59 PM
HallsofIvy
Re: Probability
If you have n distinct object, they can be arranged in n! different ways. If m of those things are the same, divide by m! to account for the fact that for each way of arranging the n things, m! of them are identical except for rearrangements of just those things. For example, there are $\displaystyle \frac{6!}{2!2!}= \frac{720}{4}= 180$ ways to arrange the letters aabcdd. Apply that to your problem. (The correct answer is NOT 120.)

There are 3 ways in which we could have Mathematics, Physics, and Chemistry on the first day and 7 ways in which not.

If any one of Mathematics, Physics, or Chemistry were given on the first day, then there must be one other 7 courses. So far the number of different ways is 3(7)= 21.
If it was one of the other courses on the first day, there are 6 ways we can give a "non math, physics, chem" test- 7(6)= 42- and 3 ways we can give one of those courses:7(3)= 21.

If it was one of math, physic, chem on the second day, it must be one of the other 6 courses: 7(3)(6)= 126

Continue like that. Add the separate possible combinations to find the total.