how many 10 digit tenary sequences contain exactly 3 1s and

how many 10 digit tenary sequences contain exactly 3 1\'s and 2 0\'s?

Solution

There are
C(10,3) = 10! / (7! 3!) = (10* 9*8) /(3* 2) = 120 possible combinations of positions for the 3 1\'s.

Of the 7 remaining positions, there are
C(7,2) = 7! / (5! 2!)
= 7*6 / ( 2 * 1)
= 21 possible possible combinations of positions for the two 0\'s.

The remaining 5 places are, of course, occupied by the five 2\'s.

So there are 120 * 21 = 2520 such strings.