# Why does the wording of how many ways can a photographer 6 people from a group of 10 ask for permutations and not combinations?

Note: Please do not post the mathematical notation for binomial coefficient or "n choose m" or anything related to that. The chapter where that is introduced comes much later. Therefore I would not have access to that knowledge directly.

I am very confused by the wording of this question from Kenneth Rosen's Discrete Mathematics 6th Edition.

p. 346 problem 40

1. In how many ways can a photographer at a wedding arrange 6 people in a row from a group of 10 people, where the bride and the groom are among these 10 people, if

a) the bride must be in the picture?

b) both the bride and the groom must be in the picture?

c) exactly one of the bride and the groom is in the picture?

This problem comes after chapter 5.1, which is an introduction to the basic principles of counting: product rule, sum rule, and inclusion-exclusion principle. The chapter never deals with permutations or combinations (these terms are not even defined yet). However, the answers to this question suggest that it is asking for permutations and not just combinations. What wording of the question says I am looking for permutations and not combinations?

For example, to solve part a), I calculated all the (unique) combinations of 6 people from a group of 10 where 1 is always the bride. This comes to $1 \times 9 \times 8 \times 7 \times 6 \times 5 = 15,120$. But the number of permutations is actually $6 \times 15120 = 90720$, since for each combination the bride has 6 possible positions. This sort of makes sense.

The same reasoning is used to solve b) (permutations again), which comes to $1 \times 1 \times 8 \times 7 \times 6 \times 5 = 1,680$ combinations times $6 * 5$ ways to permute each combination, so the answer is $30 * 1680 = 50,400$.

However, part c) is incredibly confusing because the answers I have found seem to just use combinations, and not permutations. Assuming my reasoning is correct in both a) and b), the number of combinations of 6 people from a group of 10 where either the bride or the groom (but not both) is chosen is just $1 \times 8 \times 7 \times 6 \times 5 \times 4 = 40,320$ for the bride and for the groom. However, this is just combinations, and not permutations to my understanding, because there are 6 ways to arrange either the bride or groom for each such combination. Yet, the answers I have found simply add both combinations together: $40320 + 40320 = 80,640$. If we were considering permutations, would it not need to be $[6 * 40320] + [6 * 40320] = 483,840$ permutations?

So my question is, why is the solution to part c) asking only for combinations, where as a) and b) both ask for permutations, when the scope of the question never changed (arrange still has scope over part c)).

I am confused overall by this question as well, since the author never introduced any permutations in chapter 5.1 and these are the exercises for 5.1.

• I had the exact same question. The placement of this problem in the book doesn't make complete sense. – joshmcode Nov 28 '15 at 15:46

As you noted, "arrange ... in a row" indicates that order does matter, i.e., we are counting permutations. The reasoning you used in parts (a) and (b) is correct. For part (c), one way to count the permutations is

• $2$ ways to choose exactly one of the bride and groom,
• $6$ ways to choose their position,
• $8\cdot7\cdot6\cdot5\cdot4=6720$ ways to choose and arrange the other five people.

Thus, there are $80640=(2\cdot6)\cdot8\cdot7\cdot6\cdot5\cdot4$ permutations in this case.

• Oh I see now. I am looking back in my work and this is how I did it. For some reason I thought the number of combinations was 40320 and not 6720! Thank you. – user3898238 Sep 24 '14 at 19:34

Since you are choosing from the bride or the groom, the answer is

2 x 8 x 7 x 6 x 5 x 4

not

1 x 8 x 7 x 6 x 5 x 4

One person from bride and groom can be selected by 2c1 way. Now we need 5 more persons = 8c5 there is no restriction about their position so can be arranged 6! Answer = 2c1 * 8c5 * 6! =80640