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How many 3-digit even numbers can be made using the digits 1, 2, 3, 4, 6, 7, if no digit is repeated?

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प्रश्न

How many 3-digit even numbers can be made using the digits 1, 2, 3, 4, 6, 7, if no digit is repeated?

योग
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उत्तर

3-digit even numbers are to be formed using the given six digits, 1, 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.

Number of ways of filling hundreds and tens place 

= 5P2 = `(5!)/((5 - 2)!) = (5!)/(3!)`

= `(5 xx 4 xx 3!)/(3!)`

= 20

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

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अध्याय 6: Permutations and Combinations - EXERCISE 6.3 [पृष्ठ ११४]

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एनसीईआरटी Mathematics [English] Class 11
अध्याय 6 Permutations and Combinations
EXERCISE 6.3 | Q 3. | पृष्ठ ११४

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