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How many different numbers can be formed with the digits 1, 3, 5, 7, 9, when taken all at a time, and what is their sum? |
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Answer» The total number of numbers = 5 ! = 120. Suppose we have 9 in the unit’s place. We will have 4 ! = 24 such numbers. The number of numbers in which we have 1, 3, 5 or 7 in the unit’s place is also 4 ! = 24 in each case. Hence the sum of the digits in the unit’s place in all the 120 numbers = 24 (l + 3 + 5 + 7 + 9) = 600. The number of numbers when we have any one of the given digits in ten’s place is also 41=24 in each case. Hence the sum of the digits in the ten’s place = 24 (1 + 3 + 5 + 7 + 9) tens = 600 tens = 600 × 10. Proceeding similarly, the required sum = 600 units + 600 tens + 600 hundreds + 600 thousands + 600 ten thousands = 600(1 + 10 + 100 + 1000 + 10000) = 600 × 11111 = 6666600. |
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