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June 29th, 2019, 02:00 AM   #1
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hOW TO SOLVE THIS

Hello sir,
How to solve this? The number of even numbers of three digit which can be formed with digit 0,1,2,3,4 and 5 (no digit can be used more than once) is
1. 60
2. 92
3. 52
4. 48

Last edited by skipjack; July 1st, 2019 at 02:30 PM.
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June 29th, 2019, 09:08 AM   #2
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Even number, so the third digit must be 0, 2, or 4 (three possibilities).
Second digit must be one of the five remaining digits.
First digit must be one of the four remaining.

Does that help?
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June 29th, 2019, 10:35 AM   #3
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That doesn't account for the zero in the first place.
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June 29th, 2019, 10:51 AM   #4
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Last digit is always even, i.e. 0, 2, 4

Assume first digit is even.
There are 2 ways to assign the first digit, and 2 ways to assign the last digit.
Then having assigned these there are 4 ways to assign the 2nd digit.
This is a total of 16 numbers.

Assume the first digit is odd.
There are 3 ways to assign the first digit, 3 ways to assign the last digit.
Then there are 4 ways to assign the 2nd digit.
This is a total of 36

36+16 = 52
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June 30th, 2019, 06:15 AM   #5
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digit choices ...

Odd, Odd, Even
$3 \cdot 2 \cdot 3 = 18$

Odd, Even, Even
$3 \cdot 3 \cdot 2 = 18$

Even, Odd, Even
$2 \cdot 3 \cdot 2 = 12$

Even, Even, Even
$2 \cdot 2 \cdot 1 = 4$
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July 1st, 2019, 02:41 PM   #6
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There are 5 × 4 = 20 3-digit numbers that end in 0. There are similarly 20 that end in 2 and 20 that end in 4, making 60 in total.

If, however, numbers that have a leading zero are excluded, the 8 numbers 012, 014, 024, 032, 034, 042, 052 and 054 are excluded. and so the total is 60 - 8 = 52.
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