Question

the no.of three digit even no. that xan be formed without repetition are?​

Answer 1

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Hence, the desired number is 41 × 8 = 328 numbers 3-digit even numbers exist with no repetitions. Units digit can be among 0,2,4,6,8. There are some good answers here.

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Answer 2

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☞The desired number is 41 × 8 = 328 numbers 3-digit even numbers exist with no repetitions. Units digit can be among 0,2,4,6,8.

Step-by-step explanation:

➜A three digit even number is to be formed from given 6 digits 1,2,3,4,6,7.

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HTO

Since, for the number is to be even , so ones place can be filled by 2,4 or 6. So, there are 3 ways to fill ones place.

Since, repetition is not allowed , so tens place can be filled by remaining 5 digits. So, tens place can be filled in 5 ways.

Similarly, hundred's place can be filled by remaining 4 digits. So, hundred's place can be filled in 4 ways.

So,required number of ways in which three digit even numbers can be formed from the given digits is 4×5×3=60

→Alternative Method:

3-digit even numbers are to be formed using the given six digits, ,2,3,4,6 and 7, without repeating the digits.

Then, units digits can be filled in 3 ways by any of the digits, 2,4 or 6.

Since the digits cannot be repeated in the 3-digit numbers and units place is already occupied with a digit (which is even), the hundreds and tens place is to be filled by the remaining 5 digits.

Therefore, the number of ways in which hundreds and tens place can be filled with the remaining 5 digits Is the permutation of 5 different digits taken 2 at a time.

$5p2 = \frac{5}{(5 - 2)} \: = \frac{5}{3}$

Number of ways of filling hundreds and tens place

$\frac{5 \times 4 \times 3}{3} = 20$

Thus, by multiplication principle, the required number of 3-digit numbers is 3×20=60