Question:

Step 1: Calculate the total number of ways to select 7 digits from the 9 available digits (1 through 9). The number of ways to choose 7 distinct digits from a set of 9 distinct digits is given by the binomial coefficient $\binom{9}{7}$. $$ \binom{9}{7} = \binom{9}{9-7} = \binom{9}{2} = \frac{9 \times 8}{2 \times 1} = 36 $$ Step 2: Consider dividing a specific set of 6 digits into two unordered groups of 3 digits each. The solution refers to "The digits excluding 9 (the greatest) number 6", implying a set of 6 specific digits. The initial distribution into two groups of 3, where the order of the groups does not matter, is represented as: $$ \binom{6}{3} \times \binom{6}{3} \times \frac{1}{2!} $$ Here, $\binom{6}{3}$ selects 3 digits for the first group, and the remaining $\binom{6}{3}$ (this should logically be $\binom{3}{3}$ for the remaining 3 items, but the original solution writes $\binom{6}{3}$ again) selects for the second group. The factor $\frac{1}{2!}$ accounts for the groups being unordered. Calculating the binomial coefficient $\binom{6}{3}$: $$ \binom{6}{3} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20 $$ So, this part of the expression evaluates to: $$ 20 \times 20 \times \frac{1}{2} = 400 \times \frac{1}{2} = 200 $$ Step 3: Adjust the division for ordered groups and combine with the initial selection. The solution states, "However, the 3 digits on one side can move to the other", which implies that the order of the two groups of 3 digits is now considered. This effectively removes the $\frac{1}{2!}$ factor from the previous calculation (multiplying by $2!$). The combined expression is formed by multiplying the initial selection of 7 digits by this adjusted number of ways to divide the 6 digits. The final expression given in the original solution is: $$ \binom{9}{7} \cdot \binom{6}{3} \cdot \frac{2!}{2!} = \binom{9}{7} \cdot \binom{6}{3} \cdot \binom{6}{3} $$ Substituting the calculated values: $$ 36 \times 20 \times 20 = 36 \times 400 = 14400 $$ The final answer is $14400$.