Explanation:
We observe that 0 + 1 + 2 + 3 = 6. Hence the remaining two digits must account for the sum 4. This is possible with 4 = 0 + 4 = 1 + 3 = 2 + 2. Thus we see that the digits in any such 6-digit number must be from one of the collections:
{0, 1, 2, 3, 0, 4}, {0, 1, 2, 3, 1, 3} or {0, 1, 2, 3, 2, 2}.
Consider the case in which the digits are from the collection {0, 1, 2, 3, 0, 4}. Here 0 occurs twice and 1,2,3,4 occur one each. But 0 cannot be the first digit. Hence the first digit must be one of the 1,2,3,4. Suppose we fix 1 as the first digit. Then the number of 6-digit numbers in which the remaining 5 digits are 0,0,2,3,4 is 5!/2! = 60. Same is the case with other digits: 2, 3, 4.
Thus the number of 6-digit numbers in which the digits 0,1,2,3,0,4 occur is 60 x 4 = 240.
Suppose the digits are from the collection {0,1,2,3,1,3} .The number of 6- digit numbers beginning with 1 is 5!/2! = 60. The number of those beginning with 2 is 5!/2!2! = 30. and the number of those beginning with 3 is 5!/2! = 60. Thus the total number, in this case, is 60 + 30 + 60 =150. Alternately, we can also, count it as follows: the number of 6-digit numbers one can obtain from the collection {0,1,2,3,1,3} with 0 also as a possible first digit is 6!/2!2! = 180; the number of 6-digit numbers one can obtain from the collection {0,1,2,3,1,3} in which 0 is the first digit is 5!/2!2! = 30. Thus the number of 6- digit numbers formed by the collection {0,1,2,3,1,3} ,such that no number has its first 0 is 180-30=150.
Finally look at the collection {0,1,2,3,2,2}. Here the number of 6-digit numbers in which 1 is first digit 5!/3! = 20; the number of those having 3 as the first digit is 5!/3! = 20; Thus the number of admissible 6-digit number here is 20+60+20=100.This may also be obtained using the other method of counting: 6!/3! - 5!/3! = 120 - 20 = 100;
Finally the total number of 6-digit numbers in which each of the digits 0,1,2,3 appears at least once is 240 + 150 + 100 = 490.
