Question

For this group of questions, assume repetitions are not permitted. (a) How many 3-digit numbers can be formed from the six digits 1, 3, 5, 7, 9, 0? (b) How many of these numbers are less that 500? (c) How many are even? (d) How many are odd? (e) How many are multiples of 5?

Answer 1

There are 100 3-digit numbers that can be formed from 1, 3, 5, 7, 9, 0; 60 are less than 500, 20 are even, 80 are odd, and 20 are multiples of 5.

Step-by-step explanation:

To solve the problem of forming 3-digit numbers from the digits 1, 3, 5, 7, 9, 0 without repetition:

a. The total number of possible 3-digit numbers is calculated by considering that the first digit can be any of the five non-zero digits and the second and third can be any of the remaining five and four digits, respectively. Therefore, the total is 5 * 5 * 4 = 100 numbers.

b. To determine numbers less than 500, we can only use the digits 1, 3, and 0 for the first digit, with 3 possibilities, assuming that the second and third digits can be any of the remaining five and four digits, thus making a total of 3 * 5 * 4 = 60 numbers.

c. Numbers that are even must end in 0. There are 5 options for the first digit (excluding 0) and 4 options for the second (excluding the chosen first digit and 0), totaling 5 * 4 * 1 = 20 even numbers.

d. Since all numbers not ending in 0 are odd and we already found out the even numbers, we can subtract this from the total. Thus, the total number of odd numbers would be 100 - 20 = 80.

e. Multiples of 5 must end in 5 or 0, only 5 is an option here. We have 5 choices for the first digit, 4 for the second, and 1 remains fixed, resulting in 5 * 4 = 20 that are multiples of 5.

Answer 2

a. 100

b. 40

c. 20

d. 100

e. 40

Step-by-step explanation:

The given numbers are 1, 3, 5, 7, 9, 0

If the repetitions are not allowed,

(a) Let the three digit number is xyz.

At x number of digits that can be placed = 5 [except 0]

At y we can place digits = 5

At z number of digits we can place = 4

Total number of digits that can be formed = 5×5×4 = 100

(b). Numbers formed which are less than 500.

Let the number is in the form of xyz

Then x will be less than 5 to form numbers less than 500.

So possible number of digits at x = 2 [1, 3]

Possible number of digits at y = 5

Possible number of digits at z = 4

Total numbers formed = 2×5×4 = 40

(c) Even numbers will have 0 at the place of z.

Then number of digits at x may come = 5

Number of digits at y = 4

At z only one digit may come.

Therefore, number of even numbers formed = 5×4×1

= 20 numbers

(d) Odd number means at z there should be any one out of 1, 3, 5, 7, 9.

Therefore, at z number of digits may come = 5

At x number of digits may come = 5

At y number of digits may come = 4

Total odd numbers = 5×5×4 = 100

(e) Multiple of 5 means at z there may be either 0 or 5.

So numbers having 0 or 5 at z = 2

Now digits that may come at x = 5

Digits that may come at y = 4

Total numbers may be formed = 2×5×4 = 40