# 15-344/Homework Assignment 7

This assignment is due at the tutorials on Thursday November 19. Here and everywhere, neatness counts!! You may be brilliant and you may mean just the right things, but if the teaching assistants will be having hard time deciphering your work they will give up and assume it is wrong.

Solve problems 8, 10, 15, 20, 24, 28, and 29 in section 5.3 and problems 1, 3, 16, 39, 40, 63, and 65 in section 5.4, but submit only your solutions of the underlined problems.

 Dror's notes above / Students' notes below

Homework Assignment 8 Solution

15 How many 8-digit sequences are there involving exactly six different digits?

A) We have 2 cases: 1 digit repeated three times (for example, 11123456) and 2 digits repeated (for example, 11223456)

For the first case, there are $\binom{10}{6}$ ways to choose 6 digits from 10 digit choice. Then there are $\frac{8!}{3!}$ arrangements involving a triple digit and and $\binom{6}{1}$ different triple digits possible (111,222 and so on).So, in total, we have: $\binom{10}{6} \binom{6}{1}\frac{8!}{3!}$

Similarly, for the second case, we have: $\binom{10}{6} \binom{6}{2}\frac{8!}{2!2!}$

The sum of the two cases gives the answer:

$\binom{10}{6}(\binom{6}{1}\frac{8!}{3!}+ \binom{6}{2}\frac{8!}{2!2!})$