What we see: a Problem Solving counting question. We line up 7 digits, three of them are the same (3, 3, 3), and the number cannot start with 0.
Step 1. Count every order of the 7 digits.
What we see: 7 digits, so if they were all different there would be \(7! = 5040\) orders.
What it means: the three 3s look the same. Swapping them gives the same number, and three 3s can be swapped \(3! = 6\) ways.
What we do and why: divide by 6 to stop counting the same number six times: \(\frac{7!}{3!} = \frac{5040}{6} = 840\).
Step 2. Remove the orders that start with 0.
What we see: a number like 0133378 is really the 6-digit number 133378, so it does not use every digit as a 7-digit number.
What it means: we must throw out every order with 0 in front.
What we do and why: fix 0 in the first place and arrange the other 6 digits (1, 3, 3, 3, 7, 8): \(\frac{6!}{3!} = \frac{720}{6} = 120\).
Step 3. Subtract.
\(840 - 120 = 720\).
The trap: D, 4320. That is \(7! - 6! = 5040 - 720\): the 0 is handled right, but the three 3s are counted as if they were different digits, so every number is counted 6 times.
Faster way: the first digit can be anything but 0, which is 6 of the 7 digits, so 6 out of every 7 orders are fine: \(\frac{6}{7} \times 840 = 720\).
Takeaway: repeated items: divide by the factorial of each repeat. A 0 that cannot lead: take away the orders that start with it.
Answer: B
