A combination lock requires three numbers from 1 to 25 How m

A combination lock requires three numbers from 1 to 25. How many combinations are possible? Did you use the sum rule or the product rule? (Note that although we use the term combination with locks, order matters.)

Solution

Solution.1

(Using multiplicative principle)

We have to select 3 numbers. For the first number there is no restriction. So the number of choices is 25. The 2nd number must be different from the first one. So the number of choices is 24. Finally the last number has to be different from the 2nd one. So there are 24 possibilities. Therefore the total number of acceptable combinations is 25×24×24=14400.

Solution 2: (Using inclusion exclusion principle)

Let U be the set of all combinations regardless it is acceptable or not. Then |U|=25³ = 15625.

Let A be the set of combinations whose 1st two numbers are same. And B be the set of combinations whose last 2 numbers are same. Then the set of non acceptable combinaions is precisely |AUB|.

|A|=|B|=25²

A intersection B = 25. Therefore

|AUB| = |A|+|B|-|A intersection B|

= 25² + 25² -25

= 625+625-25

=1225.

So the number of acceptable combinations is 15625 - 1225 = 14400.