Question

The total number of 10 digits sequences formed by only {0, 1, 2} where 1 should be used at least 5 times and 2 should be used exactly three times, is

Answer 1

3480

Answer 2

We have 10 positions.

1. Exactly three positions must be occupied by digit 2. This can be done in $\binom{10}{3}$ ways.

2. The remaining 7 positions are filled with digits from $\{0,1\}$. In these 7 spots, the digit 1 must appear at least 5 times. Let $x$ be the number of ones. Then, $x = 5, 6,$ or $7$.

For each case:

• $x=5$: $\binom{7}{5}$ ways.

• $x=6$: $\binom{7}{6}$ ways.

• $x=7$: $\binom{7}{7}$ ways.

Thus, the total number of sequences is

$\binom{10}{3} \left[\binom{7}{5} + \binom{7}{6} + \binom{7}{7}\right] = 120 \times (21 + 7 + 1) = 120 \times 29 = 3480$.

Explanation (Minimal):

Choose three positions for 2's: $\binom{10}{3}$. Fill the remaining 7 positions with 0's and 1's such that ones appear at least 5 times: $\binom{7}{5}+\binom{7}{6}+\binom{7}{7}=29$. Total = $120 \times 29 = 3480$.