Question

There are 6 students who all want to be in a group of 3 people chosen to speak at graduation. How many different groups can be chosen?

Answer 1

i'd say 10

Answer 2

You need to find the number of combinations of the 6 students taken 3 at a time:
\[\large 6C3=\frac{6!}{3!3!}=\frac{6\times5\times4}{3\times2\times1}=you\ can\ calculate\]

Answer 3

120/6

Answer 4

?

Answer 5

Yes, 20 is correct.

Answer 6

really? i did it by writing it. the first time going across: 1,2,3,4,5,6= 4 ways. did it again and got 3 ways. again and got 2 times. last i got 1. then i added them to get 10

Answer 7

so final answer is 20?

Answer 8

123 and 456,124 and 356, 125 and 346. then 234 and 156, 235 and 146, 236 and 145. then 345 and 126. so i got 7? confused

Answer 9

The number of combinations of n different things taken r at a time can be written nCr.\[\large nCr=\frac{n(n-1)(n-2)\ .........\ (n-r+1)}{1\times2\times3\ ...... \times r}=\frac{n!}{r!(n-r)!}\]

Answer 10

Yes, the number of combinations of 6 students taken 3 at a time is 20. Therefore 20 different groups can be chosen.

Answer 11

alright. i'd go with what he said

Answer 12

can you help me with this one?

Answer 13

you have 20 books and want to make a reading list that has 3 books on it. How many different reading lists can you make? (2 reading lists are different if they have the same books but in a different order

Answer 14

6,840 i think if i follow crept correctly

Answer 15

kropot. think my computer "corrected" it to crept

Answer 16

np! In the question with the books, the number of permutations of 20 different books taken 3 at a time is required.
\[\large 20P3=\frac{20!}{(20-3)!}=20\times19\times18\]

Answer 17

6,840? correct

Answer 18

kropot is correct with the equation which would equal to 6840