Answer to Question #329942 in Statistics and Probability for khan

Question

a) How many three-digit numbers can be formed from the digits 0, 1, 2, 3, 4, 5, and 6 if each digit

can be used only once? (05)

(b) How many of these are odd numbers? (05)

(c) How many are greater than 330?

Answer 1

For the first question, consider that you have six options for the first digit (1-6), six options for the second digit (0–6 except the number used for the first digit), and five options for the third digit (0–6 except the numbers used for the first two digits). So you can create 6$\cdot$ 6$\cdot$ 5=180

6$\cdot$ 6$\cdot$ 5=180 three-digit numbers.

To find the number of odd numbers, pick the third digit first - it can be either 1, 3, or 5, so you have three options. Then you have five options for the first digit (1–6 except the number used for the third digit) and five options for the second digit (0–6 except the numbers already used). So of the 180 numbers total, 3$\cdot$5$\cdot$ 5=75

3$\cdot$5$\cdot$ 5=75 of them are odd.

Finally For the third one greater than 330 we have two ways to achieve this either draw a 4, 5 or 6 for the first digit then we don't care about the others making 3$\cdot$ 6$\cdot$5=90

3$\cdot$6$\cdot$5=90

Alternatively we must draw a 3 for our first digit then 4, 5 or 6 for our second and finally any digit making 1$\cdot$3$\cdot$5=15

1$\cdot$3$\cdot$5=15

Now since both of these groups have no overlap we can simply add them together to get 90+15=105

90+15=105.