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Saturday, December 13, 2008

Q) 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.
(From RMO - 2008)

A) The various possibilities of the digits chosen acc. to the conditions mentioned are:

1) Three 2s, One 3, One 1, One 0
2) One 2, Two 3s, Two 1s, One 0
3) One 4, One 2, One 3, One 1, Two 0


Number of permutations for the cases mentioned above:

1) Three 2, One 3, One 1, One 0
(6! / 3!) - (5! / 3!) -----> ie. net permutations - permutations having 0 as first digit since those won't be considered a 6 digit number. Similar for other cases too.
= 5.5! / 3!
= 100 ways

2) One 2, Two 3, Two 1, One 0
(6! / 2!.2!) - (5! / 2!.2!)
= 5.5! / 4
= 150 ways

3) One 4, One 2, One 3, One 1, Two 0
(6!/2!) - 5!
= 240 ways

Total = 100 + 150 + 240
= 490 such numbers


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