**Problem 3**

A) How many four-digit whole numbers occur such that the left-most digit is odd, the second digit is even, and all four digits are different?

B) How many four-digit whole numbers have an even left-most digit, an odd second digit, and four different digits?

[Problem submitted by Iris Magee, LACC
Associate Professor of Mathematics.]

**Solution for Problem 3:**

A) 1400. Five odd and five even digits can be used for
the two left most digits in the number.
Once an odd and an even digit have been selected and since all four
digits are different, eight choices remain for the third digit, and then seven
choices for the fourth digit.

Thus 5 x 5 x 8 x 7,
or 1400, such whole numbers exist.

B) 1120. Four even (because cannot use 0 for first
digit) and 5 odd digits can be used for the two left most digits in the
number. Once an even and an odd have
been selected and since all four digits are different, 8 choices remain for the
third digit, and then 7 choices for the fourth digit.

Thus we have 4 x 5 x
8 x 7= 1120.