Question

The number of arrangements of the numbers strictly between 10 and 1000 formed with the digits 0, 1, 2, 3, 4, 5, 6 without repetition is

• A. \(180\)
• B. \(215\)
• C. \(216\)
• D. \(222\)

Hint:

Count two-digit and three-digit numbers separately, then remove 10 itself.

Answer 1

The Correct Option is B

Step 1: Count with permutations:
Three-digit: choose an ordered triple of distinct digits from 7 and subtract those with leading zero: $7P3 - 6P2 = 210 - 30 = 180$.

Step 2: Two-digit:
Two-digit: $7P2 - 6P1 = 42 - 6 = 36$.

Step 3: Adjust for the bounds:
The range is strictly above 10, so remove 10 itself: $36 - 1 = 35$. Total is $35 + 180 = 215$, option (B).

Final Answer:
Total arrangements are 215. \[ \boxed{\text{(B) }215} \]