How many five digit numbers can be formed from the set of nine numbers (0,1,2,3,4,5,6,7,8) if no number is repeated and no number starts with zero and

a) there are no other restrictins
b) the result must be an odd number

a) no restrictions, except can't start with a zero

so 8x8x7x6x5 = 13440

b) has to be odd, not start with zero
has to end with odd number, there are 4 of these
can't start with zero, leaving 7 (remember one is already gone in the odd spot)

number of cases = 7x7x6x5x4 = 5880

thank you!

To solve this problem, we can break it down into two parts:

a) First, let's consider the case where there are no other restrictions.

To form a five-digit number from the set of nine numbers (0,1,2,3,4,5,6,7,8), we need to choose 5 numbers from the set without repetition.

The number of ways to choose 5 numbers from a set of 9 without repetition is given by the combination formula:

C(n, r) = n! / (r! * (n-r)!)

Where n is the total number of elements in the set and r is the number of elements we want to choose. In this case, n = 9 and r = 5.

C(9, 5) = 9! / (5! * (9-5)! )
= 9! / (5! * 4! )
= (9 * 8 * 7 * 6 * 5!) / (5! * 4 * 3 * 2 * 1)
= 9 * 8 * 7 * 6
= 3,024

Therefore, there are 3,024 different five-digit numbers that can be formed from the set of nine numbers (0,1,2,3,4,5,6,7,8) if no number is repeated and no number starts with zero.

b) Now, let's consider the case where the result must be an odd number.

If we want to form a five-digit number that is odd, the last digit must be odd. Out of the given set (0,1,2,3,4,5,6,7,8), there are four odd digits (1, 3, 5, 7) and four even digits (0, 2, 4, 6).

The last digit can be chosen in 4 ways (either 1, 3, 5, or 7).

For the remaining four digits, we need to choose 4 digits without repetition. Since the first digit cannot be zero, we have 8 choices (excluding zero), and for the remaining three digits, we have 7, 6, and 5 choices respectively.

Therefore, the total number of five-digit numbers that can be formed from the set of nine numbers (0,1,2,3,4,5,6,7,8) if no number is repeated, no number starts with zero, and the result must be an odd number is:

4 (ways to choose the last digit) * 8 (ways to choose the first digit) * 7 (ways to choose the second digit) * 6 (ways to choose the third digit) * 5 (ways to choose the fourth digit)
= 4 * 8 * 7 * 6 * 5
= 6,720

Therefore, there are 6,720 different five-digit numbers that can be formed from the set of nine numbers (0,1,2,3,4,5,6,7,8) if no number is repeated, no number starts with zero, and the result must be an odd number.