CMA Foundation · Fundamentals of Business Mathematics and Statistics · Permutation and Combinations
How many 3-digit numbers with all digits different can be formed using the digits 1, 2, 3, 4, 5, 6 so that the number is even?
There are 60 such numbers. The units place must be even, giving 3 choices (2, 4, 6). The hundreds place can then be filled in 5 ways and the tens place in 4 ways, so 3 × 5 × 4 = 60.
- A60Correct
- B90
- C120
- D36
Explanation
The units digit must be 2, 4 or 6: 3 choices. The hundreds digit then has 5 choices and the tens digit has 4 choices. Total = 3 × 5 × 4 = 60. The value 120 is 6P3, which counts all numbers, not only even ones.
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