i dont understand this math problem.....?

Question

i dont understand this question: how many different 4-person relay teams can be chosen from 8 students? i have to list all of the combinations


thank u in advance, this is hard for me.

none of the combinations can be the same.

Answer 1

Let the 8 students be A,B,C,D,E,F,G,and H
Start with A as the first person
ABCD is possibility 1
ABCE is possibility 2
ABCF is possibility 3
ABCG
ABCH
ACDE
ACDF
ACDG
ACDH
ADEF
ADEG
ADEH
AEFG
AEFH
AFGH
Now do the same with B as the 1st letter. Be careful not to include combinations that are the same. ABCD is same as ADBC and ACBD etc.

Happy writing

Answer 2

1 2 3 4, 1235, 1236, 1237 1238
1245 1246 1247 1248
1256 1257 1258
1267 1268
1278
1345, 1346, 1347, 1348
1356 1357 1358
1367 1368
1378
1456, 1457, 1458,
1467 1468
1478
1567, 1568
1578
1678,
2345, 2346, 2347,2348
2356 2357 2358
2367 2368
2378
2456 2457 2458
2467 2468
2478
2567 2568
2578
2678
3456.3457.3458
3467 3468
3478
3567 3568
3578
3678
4567, 4568
4678
5678

(8x4) squared 1,024 teams
EDIT: I just scratched the surface on teams here, follow the pattern, i goofed on my first answer, sorry
8 times is 32, 32 squared is 1024
that's the answer
EDIT2: I went to mathforum.org and got a diff answer
it's 4 to the 8th power, 65536
or 8 to the 4th power 4096
I'm confused

Answer 3

The topic is COMBINATIONS. You want to see how many combinations of 4 people you can get out of 8. In simple math terms, you want to find 8C4. Typing that in your scientific calculator will give you the answer.

Here's how you do the working.
Number of possible 4-person teams out of 8 students = 8C4
8C4 = (8x7x6x5)/(4x3x2x1)
= 1680/24
= 70 combinations.

Answer 4

For your question,
if you select 4 persons from 8 students but not allow interchnage of position in relaty teams, like 1234 not same as 2134
then you can use
C(8,4) = 8!/4! = 8x7x6x5 = 1680 combinations

However,
if you select 4 persons from 8 students but allow interchange of position in relay teams, like 1234 same as 2134
then you can use
C(8,4)/4! = (8!/4!)/4! = 8x7x6x5/4x3x2x1 = 70 combinations

Answer 5

This is a problem of combination and the problem may be solved in the following way
The no of combination of 8 different things taken 4 at a time
=8!/4!(8-4)! (the sign indicates permutation
=(8x7x6x5x4x3x2x1)/4x3x2x1x4x3x2x1
=(8x7x6x5)/(4x3x2)
=70

Answer 6

Call them a, b, c, d ,e, f, g, h
Make a box 8 across and 8 down
Now go across and choose 3 letters under each of the top letters in each column.
See how many of these are different from each other.
That is your answer.

Answer 7

Ok, think of it this way:

There are 8 boys Adam, Bob, Chris, Dave, Elmo, Frank, George and Henry.

Start making a list

Adam, Bob, Chris, Dave
Adam , Bob, Chris, Elmo....

Bob, Chris, Dave, Elmo
Bob, Dave, Elmo, Frank.....

After you make the entire list, count how many teams you have. How could you have done it using a formula? How many students? How many teams? How did you get that answer?

Answer 8

I think the answer is 4 and 4 go from there

Answer 9

FIRST LABLE EACH STUDENT AS STUDENT 1 -8. LET S1= STUDENT 1 S2= STUDENT 2 ETC.
S1 S2 S3 S4 S5 S6 S7 S8

NOW COME UP WITH THE TEAMS OF 4 LIKE A PATTERN START WITH THE EASIEST COMBINATION... THE FIRST 4 STUDENTS AS A TEAM. THEN CHANGE THE LAST STUDENT IN EACH TEAM. CREATING DIFFERENT POSSIBILITY OF TEAMS. WHEN YOU HAVE ALL TEAMS INVOLVING STUDENT ONE START WITH STUDENT 2 AND SEE HOW MANY DIFFERENT TEAMS CAN INVOLVE STUDENT 2 WHICH DOES NOT INCLUDE STUDENT ONE. JUST REMEMBER TO LOOK FOR THE PATTERN FORMING:
S1 S2 S3 S4
S1 S2 S3 S5
S1 S2 S3 S6
S1 S2 S3 S7
S1 S2 S3 S8

S1 S3 S4 S5
S1 S3 S4 S6
S1 S3 S4 S7
S1 S3 S4 S8

S1 S4 S5 S6
S1 S4 S5 S7
S1 S4 S5 S8

S1 S5 S6 S7
S1 S5 S6 S8

S1 S6 S7 S8

--------------------
S2 S3 S4 S5
S2 S3 S4 S6
S2 S3 S4 S7
S2 S3 S4 S8

S2 S4 S5 S6
S2 S4 S5 S7
S2 S4 S5 S8

S2 S5 S6 S7
S2 S5 S6 S8

S2 S6 S7 S8
-------------------

S3 S4 S5 S6
S3 S4 S5 S7
S3 S4 S5 S8

S3 S5 S6 S7
S3 S5 S6 S8

S3 S6 S7 S8
------------------

S4 S5 S6 S7
S4 S5 S6 S8

S4 S6 S7 S8
------------------

S5 S6 S7 S8

NOW COUNT THE POSSIBILITYS CREATED: 35

Answer 10

35 just make number 1234 5678 2345 6781 3456 7812 you go on its simple you should only get 35