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How many even numbers between 100 and 1000 can be formed with the digits 3, 5, 6, 7, 9, if the repetition of digits is not allowed?

How many even numbers between 100 and 1000 can be formed with the digits 3, 5, 6, 7, 9, if the repetition of digits is not allowed? Correct Answer 12

Fundamental principal of multiplication:

Let us suppose there are two tasks A and B such that the task A can be done in m different ways following which the second task B can be done in n different ways. Then the number of ways to complete the task A and B in succession respectively is given by: m × n ways

Fundamental principal of addition:

Let us suppose there are two tasks A and B such that the task A can be done in m different ways and task B can be completed in n ways. Then the number of ways to complete either of the two tasks is given by: (m + n) ways.

Calculation:

Here, we have to form a 3 digit number using the digits 3, 5, 6, 7, 9 such that repetition of digits is allowed.

In order to form a 3 digit even number, units digit should be occupied by the digits like 0, 2, 4, 6 or 8.

So, unit’s digit can be filled with only 6 from the given digits.

The number of ways to fill the unit’s digit = 1

The ten’s digit can be filled with any given digits except the one present at units digit

⇒ No. of ways to fill ten’s digit = 4

The hundredth digit can be filled with any given digits except the one present at units and tens digit

⇒ No. of ways to fill ten’s digit = 3

So, number of such even numbers that can formed lying between 100 and 1000 = 4 × 3 × 1 = 12

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