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Determine the number of six-digit integers (no leading zeros) in which (a) no digit may be repeated; (b) digits may be repeated. Answer parts (a) and (b) with the extra condition that the six-digit integer is (i) even; (ii) divisible by 5 ; (iii) divisible by 5____

User Kavko
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Final answer:

In this question, we determine the number of six-digit integers with no repeated digits and with repeated digits. We also consider the extra condition of the integers being even or divisible by 5.

Step-by-step explanation:

In part (a), the number of six-digit integers with no repeated digits can be found using the concept of permutations. Since there are no leading zeros, the first digit can be chosen from 9 options (1 to 9), the second digit can be chosen from 9 options (any digit except the one already chosen), and so on. Therefore, the number of such integers is 9 * 9 * 8 * 7 * 6 * 5 = 136,080.

In part (b), the number of six-digit integers with repeated digits can be found by using the concept of combinations. Each digit can be chosen from 10 options (0 to 9), and there are 6 digits. Therefore, the number of such integers is 10^6 = 1,000,000.

In both parts (a) and (b), if the six-digit integer needs to be even, the last digit can only be chosen from the set {0, 2, 4, 6, 8}. Therefore, the number of even six-digit integers with no repeated digits is 9 * 9 * 8 * 7 * 6 * 5 * 5 = 68,400, and the number of even six-digit integers with repeated digits is 10^5 * 5 = 500,000.

If the six-digit integer needs to be divisible by 5, the last digit can only be 0 or 5. Therefore, the number of six-digit integers divisible by 5 with no repeated digits is 9 * 9 * 8 * 7 * 6 * 2 = 48,384, and the number of six-digit integers divisible by 5 with repeated digits is 10^5 * 2 = 200,000.

User Sabrehagen
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