Question

In Dr. Mosley's waiting room, 4 magazines are scattered on the chairs. A patient named Tristan decides he probably has time to skim 3 magazines before his name is called. In how many orders can Tristan pick out 3 of the 4 magazines?

Answer 1

Answer: Tristan can pick them out in 6 different ways

Step-by-step explanation: In the waiting room the four magazines all have an equal chance of being picked first and then others would be picked subsequently. If we are to pick magazine A first of all, then the others would be picked as B, C and D, or C, B and D, or D, B and C, and so on.

However, rather than spend so much time counting the different ways we can apply the mathematical method of permutation. Since choosing the first one means we can’t choose it again but others have to be chosen, and all four magazines each has an equal chance of being chosen first, then the number of all possible permutations is given as 4! (four factorial).

The question requires us to chose three out of the four magazines, so we shall apply 3!.

3! = 3 x 2 x 1

3! = 6

Therefore, there are 6 different ways to pick three out of the four magazines

Answer 2

Answer:

24 ways

Step-by-step explanation:

Combination and permutations are quite alike but they still differ in several ways.

Just as combinations is only interested in arrangements, permutations wants both arrangements and order,but I'm going to give a proper definition of permutations since the question refers to order.

In mathematics, a permutation of a set is, loosely speaking, an

arrangement of its members into a sequence or linear order, or if

the set is already ordered, a rearrangement of its elements.

Permutations is assigned the formula

P= n!/(n - k)!

And since 4 newspapers are available to the patient and there is a need to know the order with which he skims three of those magazine, then we have

P= 4!/(4 - 3)!

= 4!/(1!)

= 4!

= 4 × 3 × 2 × 1

= 24 ways