numbers and numbers.....
Number Theory
Level
3
find the number of all 6 digit natural numbers such that the sum of their digits is10 and each of the digits 1,2,3,0 occur at least once in them.
The answer is 490.
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1 solution
0+1+2+3=6 so the remaining 2 digits must add up to 4 making them either 0 and 4, 1 and 3 or 2 and 2. With 0 and 4 we are looking for the number of different orders for the digits 0, 0, 1, 2, 3 and 4. 0 cannot go at the beginning to so this is4*5!/2!=240
Next we need the possible combinations of 0, 1, 1, 2, 3 and 3. Again we cannot start with 0 so the number is 5!/2!+5!/2!+5!/2!*2!=150
Next is the combinations for 0, 1, 2, 2, 2, 3 which will be 5!/3!+5!/3!+5!/2!=100
Adding these numbers gives 490 which should be the right answer.