Find the number of all 6-digit natural numbers such that the sum of their digits is 10 and each of the digits 0,1,2,3 occurs at least once in them.
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5 Answers
620,1,2,3 occur atleast once..
now they in themselves have a sum of 6
sum of the digits is 10
so the remaining 2 digits need to make up for 4 more
so the ways are
0 and 4
1 and 3
2 and 2
Find the sum with these three cases :)
done :)
1I did it.. I need to know the correct answer .. I got 490 but many said many different answers lol
62case 1 .. . 0,0,1,2,3,4
So the total no of ways is 4.5!/2!
case 2... 0,1,1,2,3,3
first digit 1: 5!/2!
first digit 2: 5!/(2!.2!)
first digit 3: 5!/2!
case 3... 0,1,2,2,2,3
first digit 1: 5!/3!
first digit 2: 5!/2!
first digit 3: 5!/3!
so the answer is 490.. i made a slight calc mistake in taking 3! as 3 :P