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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## Saturday, December 13, 2008

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