# Math Help - Some more factorial word problems

1. ## Some more factorial word problems

I find these a bit more complicated from the problem in my previous post. I can't seem to apply that logic to these problems.

1. How many 3-digit positive integers are odd and do not contain the digit "5"?

2. A talent contest has 8 contestants. Judges must award prizes for first, second, and third places. If there are no ties, in how many different ways can the 3 prizes be awarded?

I'm sure these are so simple, but I've tried many different approaches and I keep getting wrong answers. (I'm studying for a standardized test...)

Thank you.

2. Originally Posted by alaskamath
I find these a bit more complicated from the problem in my previous post. I can't seem to apply that logic to these problems.

1. How many 3-digit positive integers are odd and do not contain the digit "5"?
Again, imagine yourself selecting digits so as to get a number $d_3 d_2 d_1$ that satisfies the given requirements (where $d_{1,2,3}$ are the digits).
In order to make sure that the number will be odd, you must choose an odd digit for $d_1$. There are 5 odd digits, but since the digit 5 must not occur anywhere in the number you are about to construct you have only 5-1=4 possibilities for $d_1$.
For $d_2$ there is no restriction placed on the digit, except that it must not be 5. So there are 10-1=9 possibilities for your choice of $d_2$.
Now for digit $d_3$: again, it must not be 5 but also, since it is the leading digit of your number, you must not make it 0 either. So there are 10-1-1=8 possible choices for $d_3$.
Since these choices are independent of each other, the total number of possibilities is the product $8\cdot 9\cdot 4$

2. A talent contest has 8 contestants. Judges must award prizes for first, second, and third places. If there are no ties, in how many different ways can the 3 prizes be awarded?
Again, imagine yourself selecting the 3 highest placed persons by yourself. For the first place, you have 8 possibilities. For each of these 8 possibilities you have 8-1=7 possibilities to choose a person for the second place. And, finally, you have 8-2=6 possibilities to choose a person for the third place. This gives $8\cdot 7\cdot 6$ different ways in all.