Seven-digit telephone numbers are formed with all digits distinct. If the leftmost (first) digit is 6 and the rightmost (last) digit is 5, how many such telephone numbers are possible?

Introduction / Context:This is a direct permutations problem with positional constraints on the first and last digits and “all digits distinct.”

Given Data / Assumptions:

• Digits available: 0–9.
• Positions: D1 D2 D3 D4 D5 D6 D7.
• D1 fixed as 6; D7 fixed as 5.
• All digits in the 7-number must be distinct.

Concept / Approach:After fixing D1 and D7, fill the five middle positions with distinct digits chosen and arranged from the remaining 8 digits.

Step-by-Step Solution:Remaining pool after using 6 and 5: 8 digits.Number of ways to choose and arrange D2–D6: P(8,5) = 8 × 7 × 6 × 5 × 4 = 6720.

Verification / Alternative check:Equivalent to placing any permutation of 5 distinct digits into the middle 5 slots with order mattering.

Why Other Options Are Wrong:120 is 5! (ignores choice of which digits); 30240 and 5040 come from other permutation counts not matching constraints; 100000 ignores “all distinct.”

Common Pitfalls:Forgetting that 0 is allowed in middle positions and that digits cannot repeat.

Final Answer:6720