Problem Solving

If the first digit cannot be a 0 or a 5, how many five-digit odd numbers are there?

A. 42,500
B. 37,500
C. 45,000
D. 40,000
E. 50,000

Soln: This problem can be solved with the Multiplication Principle. The Multiplication Principle tells us that the number of ways independent events can occur together can be determined by multiplying together the number of possible outcomes for each event.

There are 8 possibilities for the first digit (1, 2, 3, 4, 6, 7, 8, 9).
There are 10 possibilities for the second digit (0, 1, 2, 3, 4, 5, 6, 7, 8, 9)
There are 10 possibilities for the third digit (0, 1, 2, 3, 4, 5, 6, 7, 8, 9)
There are 10 possibilities for the fourth digit (0, 1, 2, 3, 4, 5, 6, 7, 8, 9)
There are 5 possibilities for the fifth digit (1, 3, 5, 7, 9)

Using the Multiplication Principle:

= 8 * 10 * 10 * 10 * 5
= 40,000
(D).


I understand the solution and problem. However, I guess I don't completely understand it because I used 8!10!10!10!5! instead of the Multiplication Principle stated above.

My reasoning was that there are x! number of ways to arrange each possible digit.

Please guide! Thank you very much.

I used 8!10!10!10!5! instead of the Multiplication Principle stated above.

If you are given 8 digits and asked to form all possible 8 digit numbers. Then you say 8! ways.
BUT
If you are given 8 digits and asked to pick any one out of it. Then you have only 8 ways . You are given five such groups and asked to frame a 5 digit number. so you multiply the length of all groups.

Hope you are clear

Thank you, Jeevan!

jeevan.Gk wrote:

I used 8!10!10!10!5! instead of the Multiplication Principle stated above.

If you are given 8 digits and asked to form all possible 8 digit numbers. Then you say 8! ways.
BUT
If you are given 8 digits and asked to pick any one out of it. Then you have only 8 ways . You are given five such groups and asked to frame a 5 digit number. so you multiply the length of all groups.

Hope you are clear

Exactly - when looking at a string of numbers, we'd use permutations. When looking at each individual digit, it becomes a combinations issue.

Here's another way you could think about it:

1st digit.. pool of 8 numbers, choosing 1... 8C1
2nd-4th digits... pool of 10 numbers, choosing 1.. 10C1 (each)
last digit... pool of 5 digits, choosing 1... 5C1

So, we're multiplying:

8C1 * 10C1 * 10C1 * 10C1 * 5C1 = 8*10*10*10*5

which is really just the multiplication principle explained a different way (that can be applied to more complicated questions as well, i.e. when we're choosing more than 1 member from each group).