Problem Variations

In this scenario, there cannot be repeating values. If Bob is sitting in the first seat, he can’t also sit in the second seat. So this is a permutation without replacement.

Method 1

1. Figure out how many places there are to fill.

There are 5 seats on the bench:  ___  ___  ___  ___  ___

2. Figure out how many objects can potentially go into each place.

The number of available people decreases as people are selected:    5   4   3   2   1

3. Multiply.

5 × 4 × 3 × 2 × 1   =  5!  = 120

Method 2

5 P 5 = \dfrac{5!}{(5 \,-\, 5)!} = \dfrac{5!}{0!} = 5! = 120

Notice that 0! = 1.

This may appear to be different from the bench example, but it is conceptually identical. John cannot be served the same meal twice before trying all the other meals, so he has four days of decreasing possibilities. It is another non-replacement permutation.

Method 1

1. Figure out how many places there are to fill.

John has four days of meals:  ___  ___  ___  ___

2. Figure out how many objects can potentially go into each place.

On the first day there are 6 meals available, and the number decreases as meals are selected:
   6   5   4   3

3. Multiply.

6 × 5 × 4 × 3   =  30 × 12   = 360

Method 2

6 P 4 = \dfrac{6!}{(6 – 4)!} = \dfrac{6!}{2!}

= \dfrac{6 × 5 × 4 × 3 × 2!}{2!}

= 6 × 5 × 4 × 3 = 360

This is a non-replacement permutation.

Method 1

1. Figure out how many places there are to fill.

There are six cages for the animals, so there are six places:
  ___  ___  ___  ___  ___  ___

2. Figure out how many objects can potentially go into each place.

This step is where things change slightly. Visualize the restrictions. There are 3 cats and 3 dogs, but the cats must be in the second, fourth, and sixth cages. That means the dogs must be in the first, third, and fifth cages:    3   3   2   2   1   1

3. Multiply.

3 × 3 × 2 × 2 × 1 × 1   =  36

Method 2

This question has 2 permutations: one for the dogs and one for the cats. Find each permutation then multiply.
There are 3 dogs and 3 cats, so each has the same formula:

3 P 3 = \dfrac{3!}{(3 \,-\, 3)!} = 3! = 6

So there are 6 order possibilities for the dogs and 6 for the cats.

The total number of permutations is 6 × 6 = 36.

There are 6 letters. The letters D and O repeat.

\dfrac{n!}{\textit{s}_{\displaystyle{1}}! × \textit{s}_{\displaystyle{2}}!} = \dfrac{6!}{2! × 2!}

= \dfrac{6 × 5 × 4 × 3 × 2 × 1}{2 × 1 × 2 × 1} = 180

Note: Permutations of letters are called anagrams if the new order of the letters is also a word. For example, DAD is an anagram of ADD; and SILENT is an anagram of LISTEN. But when counting permutations of letters, each permutation does not need to form a word.

There are 6 pots. There are 4 pots that have a different plant (plant A, plant B, plant C, plant D). But the 2 pots without a plant are “repeats” (no plant, no plant).

\dfrac{n!}{\textit{s}!} = \dfrac{6!}{2!} = \dfrac{6 × 5 × 4 × 3 × 2 × 1}{2 × 1}

= 6 × 5 × 4 × 3 = 360

This question has 2 permutations: one for senior-level ID numbers and one for junior-level ID numbers. Find each permutation separately, and then make a ratio of the totals.

For senior-level employees:

1. Figure out how many places there are to fill.

There are 4 digits:  ___  ___  ___  ___

2. Figure out how many objects can potentially go into each place.

The first digit can be any digit except 0. That means there are 9 options.
The second one can have any digit including 0, but it cannot have the same digit as the one before it, so there are also 9 options.
With two digits used, there are 8 options for the third digit, then 7 options for the fourth digit.
9   9   8   7

3. Multiply.

9 × 9 × 8 × 7

But don’t complete the calculation yet. As always, keep in factored form and watch for cancellation.

For junior-level employees:

The logic is the same here, except there are 5 digits. The final step will be 9 × 9 × 8 × 7 × 6.

Finally, the question wants the ratio of these two permutations.

All the numbers cancel out except for 6. The result is simply \dfrac{1}{6}.