How many 6 digit telephone numbers can be formed if each number starts with 35…

Concept: When items must be arranged in order without repetition, the number of ways to fill k positions from n available choices is n × (n-1) × (n-2) × ... down to (n-k+1) — one fewer choice at each successive position because a digit already used cannot be used again.

1. The telephone number has 6 digits in total, and the first two digits are fixed as 3 and 5.

2. That leaves 4 positions still to be filled, chosen from the digits 0–9 excluding the two digits already used — so 8 digits remain available.

3. Since no digit may repeat, each successive position has one fewer available choice than the position before it: 8 choices for the third digit, 7 for the fourth, 6 for the fifth, 5 for the sixth.

4. Multiplying these successive counts together: 8 × 7 × 6 × 5 = 1680.

Cross-check: using the permutation formula P(8,4) = 8! / (8-4)! = 8 × 7 × 6 × 5 = 1680, which matches the count obtained directly from the multiplication principle above.

So 1680 distinct 6-digit telephone numbers can be formed under the given rule.