Question

A teacher is trying to pick 6 students to represent his class at an upcoming state competition. If there are 13 boys and 15 girls in the class, in how many ways can the teacher pick 2 boys and 4 girls to attend the event?

Answer 1

for this i thought you would do 13*12 + 15*14*13*12
but that's not right

Answer 2

@antoni7

Answer 3

You have 13 C 2 ways to pick 2 boys

You have 15 C 4 ways to pick 4 girls

Answer 4

Evaluate each piece, then multiply the two results

Answer 5

5110560?

Answer 6

that's way too big

Answer 7

ok so i divide that by something?

Answer 8

13 C 2 = ??

Answer 9

(2*1) * (4*3*2*1) ??

Answer 10

13 C 2 = ??

Answer 11

what does 13 c 2 mean? 13*12?

Answer 12

13 C 2 means you plug n = 13 and r = 2 into


(n!)/(r!*(n-r)!)

Answer 13

it's basically the number of ways to pick 2 items (out of 13) where order doesn't matter

Answer 14

ok so 78

Answer 15

how about 15 C 4

Answer 16

1365

Answer 17

multiply 78 and 1365 to get your final answer

Answer 18

ok i got it...

can you help me with this problem...

A dinner party has 6 couples. The hostess wishes to seat all the guests around a circular table alternating males and females. How many differnt seating arrangements are there? Assume that there is no special seat around the table.

Answer 19

i tried doing 12! / 6!

Answer 20

also just 12! and just 6!

Answer 21

it has 6 couples, so 6 males and 6 females

so 6! * 6!

Answer 22

1440?

Answer 23

yes

Answer 24

thats not it

Answer 25

hmm let me think then

Answer 26

its gonna be a huge number

Answer 27

what makes you say that

Answer 28

i have an example one and the answer is over 200 million. it doesn't show work though.

Answer 29

what's the example

Answer 30

A dinner party has 8 couples. The hostess wishes to seat all the guests around a circular table alternating males and females. How many differnt seating arrangements are there? Assume that there is no special seat around the table.

ANSWER: 203,212,800

Answer 31

oh i guess it's as simple as 7!*8! for the example

Answer 32

i think the first slot is taken for the first person in the couple, so that may explain why it's 7! instead of 8!

Answer 33

ah i see

Answer 34

can you help me with one final one?

Answer 35

ok

Answer 36

whats the question

Answer 37

How many ways can 6 knights out of a group of 11 knights be chosen and seated at a round table?

Answer 38

i was thinking 11*10*9*7*6*5

Answer 39

There are...

11 choices for the first seat
10 choices for the second seat
9 choices for the third seat
8 choices for the fourth seat
7 choices for the fifth seat
6 choices for the sixth seat


So there are 11*10*9*8*7*6 = 332,640 different ways to do this

Answer 40

thats what i meant. i skipped the 8 for some reason lol

Answer 41

thats wrong though...

Answer 42

hmm maybe they remove another slot like last time

Answer 43

try

11*10*9*8*7

Answer 44

yeah they do...

thanks for the help!!!
:)

Answer 45

ok great, yw