# Thread: Counting, Permutation, and Combinations

1. ## Counting, Permutation, and Combinations

Guys I have several questions that I would like to ask about:

1. In how many ways can a photographer at a wedding arrange 6 people in a row from a group of 10 people, if the bride and the groom are among these 10 people, if:
a. the bride must be in the picture
b. exactly one of the bride and the groom is in the picture

2. How many ways are there for 10 women and six men to stand in a line so that no two men stand next to each other?

3. A professor writes 40 discrete mathematics true/false questions. Of these statements in these questions, 17 are true. If the questions can be positioned in any order, how many different answer keys are possible.

1a. because the bride is already included among 6 people, therefore there are 5 more people that we can choose and arrange from, so it's 5!. I am not sure if this is right
1b. I don't quite understand the question

2. P(10,10) x P(11,6)

3. C(40,17)

I am not quite sure of all these answers, but suggestions/help to my questions are highly appreciated

2. a. the bride must be in the picture
Let us divide this problem into cases.
Code:
BXXXXX
XBXXXX
XXBXXX
XXXBXX
XXXXBX
XXXXXB
Where "X" stands for any other person. And "B" stands for bride. We will find the number of such cases and then add all of them up together.

In the first case there are $\displaystyle (1)(9)(8)(7)(6)(5) = 15120$. Similarly in each case we have the same number of arrangements. So in total we have $\displaystyle 6\cdot 15120 = 90720$.

b. exactly one of the bride and the groom is in the picture
Again divide this into cases. Case 1 is that bride is in picture and groom is not. Case 2 is that bride is not in picture and groom is. Then add those results together. Each subcase is similar to the problem just did above.

3. Originally Posted by ThePerfectHacker
Let us divide this problem into cases.
Code:
BXXXXX
XBXXXX
XXBXXX
XXXBXX
XXXXBX
XXXXXB
Where "X" stands for any other person. And "B" stands for bride. We will find the number of such cases and then add all of them up together.

In the first case there are $\displaystyle (1)(9)(8)(7)(6)(5) = 15120$. Similarly in each case we have the same number of arrangements. So in total we have $\displaystyle 6\cdot 15120 = 90720$.

Again divide this into cases. Case 1 is that bride is in picture and groom is not. Case 2 is that bride is not in picture and groom is. Then add those results together. Each subcase is similar to the problem just did above.
as for case 1:

the bride is in the picture but the groom is not, 1 x 8 x 7 x 6 x 5 x 4 = 6,720. 6720 x 6 = 13,440. I choose 8 because therefore the number of possibilities of people chosen is now 9 instead of 10 because the groom is excluded here.

and for case 2:

the bride is not in picture but the groom is has the same possibilities as case 1 right?

then we just add case 1 and case 2?

4. Originally Posted by TheRekz
as for case 1:

the bride is in the picture but the groom is not, 1 x 8 x 7 x 6 x 5 x 4 = 6,720. 6720 x 6 = 13,440. I choose 8 because therefore the number of possibilities of people chosen is now 9 instead of 10 because the groom is excluded here.

and for case 2:

the bride is not in picture but the groom is has the same possibilities as case 1 right?

then we just add case 1 and case 2?
I agree with you.

5. what if now the question is both the bride and the groom must be in the picture:

(1 x 1 x 8 x 7 x 6 x 5) = 1,680 x 6 = 10,080

is this right?

6. Hello, TheRekz!

1. In how many ways can a photographer at a wedding
arrange 6 people in a row from a group of 10 people,
if the bride and the groom are among these 10 people, and:
a. the bride must be in the picture
b. exactly one of the bride and the groom is in the picture

a) The bride is in the picture.
Select five more from the other nine people: .$\displaystyle C(9,5)$ ways.
These six people can be arranged in $\displaystyle 6!$ ways.

Therefore, there are: .$\displaystyle C(9,5) \times 6!$ ways.

b) Either bride or the groom is in the picture (but not both).
There are 2 choices for the newlywed to be in the picture.
The other 5 people are chosen from the remaining 8 people (not the other newlywed).
. . There are: .$\displaystyle C(8,5)$ ways.
The six people can be arranged in $\displaystyle 6!$ ways.

Therefore, there are: .$\displaystyle 2 \times C(8,5) \times 6!$ ways.

2. How many ways are there for 10 women and six men to stand in a line
so that no two men stand next to each other?

Arrange the tenwomen in a row. .There are $\displaystyle P(10,10) = 10!$ possible arrangements.

Leave spaces between the women: .$\displaystyle \_\;W\;\_\;W\;\_\;W\;\_\;W\;\_\;W\;\_\;W\;\_\;W\;\ _\;W\;\_\;W\;\_\;W\;\_$

Arrange the six men in any six of the eveleven spaces.
. . There are: .$\displaystyle P(11,6)$ ways.

Therefore, there are: .$\displaystyle P(10,10) \times P(11,6)$ ways.

3. A professor writes 40 discrete mathematics true/false questions.
Of these statements in these questions, 17 are true.
If the questions can be positioned in any order,
how many different answer keys are possible?

Your answer is correct . . . $\displaystyle C(40,17)$

7. Originally Posted by TheRekz
what if now the question is both the bride and the groom must be in the picture:

(1 x 1 x 8 x 7 x 6 x 5) = 1,680 x 6 = 10,080

is this right?
If both the bride and groom are in the picture then:

fix the bride in the first position:

BGXXXX = 1*1*8*7*6*5
BXGXXX = 1*8*1*7*6*5
.
.
.
BXXXXG = 1*8*7*6*5*1

Then there are 5*(1*1*8*7*6*5) ways to arrange them with the bride in the first position.

Now repeat with the bride in the second, third,....,sixth position and you get the same result each time. Therefore, if you add these all up you get (5*(1*1*8*7*6*5))*6 = 50,400

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# A professor writes 40 discrete mathematics true/false

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