KA-0140The multiplication principle
5 points, difficulty 3 of 3Level 3–4about 90sA code uses the letters A, B and C, each exactly once. How many codes do not start with A?
Hints
Take them one at a time. The first gives nothing away.
1A nudge
Count them all, then take away the bad ones.
2The strategy
Work out how many codes there are with no restriction, then how many break the rule by starting with A, and subtract. That is fewer cases than counting the good ones directly.
3The full solution
All arrangements: 3 x 2 x 1 = 6. Those starting with A: the other two letters can go 2 ways. So 6 - 2 = 4.
Solution
The reliable way
Count everything first: three different letters arrange in 3 x 2 x 1 = 6 ways. Now count the ones that break the rule: if A is fixed at the front, the remaining two letters can be arranged in 2 ways, so 2 codes start with A. That leaves 6 - 2 = 4. Listing confirms it: BAC, BCA, CAB, CBA. The transferable idea: with a restriction, counting everything and subtracting the bad cases is usually fewer steps than counting the good ones.
The elegant way
The first letter must be B or C, and the remaining two can go either way round: 2 x 2 = 4.
Why this is on the test: It is the smallest setting where complementary counting genuinely beats direct counting, so the habit forms before the numbers get large.