How many four-digit whole numbers are there such that the leftmost digit is odd, the second digit is even, and all four digits are different?

There are 1400 different combinations.

Step-by-step explanation:

The conditions are:

we have 4-digits: abcd.

all the digits are different.

a is an odd number, and b is an even number.

Then, for a, we have the options 1, 3, 5, 7 and 9 (so we have 5 options).

for b, we have the options 0, 2, 4, 6 and 8 (so we have 5 options).

for c, we can have odd or even numbers, so we have 8 options ( remember that there where 2 numbers already taken away, this is why we have only 8 options).

for d we have 7 options (because 3 numbers are already taken).

Then the number of combinations is equal to the product of the number of options for each selection:

C = 5*5*8*7 = 1400

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step-by-step explanation:

the answer is 48m^2!

step-by-step explanation: