Question

4. How many ways can 11 Redcoat basketball players and Coach Woodburn (12 people) be arranged along the sidelines:
a) with no restrictions.
b) if Coach Woodburn must be at one end of the bench?
c) if Travis and Pat must stand together?
d) if Travis, Pat, and Woody must stand together?
e) if Travis and Pat must be kept separated?

Still checking answers: as follows

Answer 1

a) 12i = 479 001 600
b) 399 16800
c) 79833600
d) " "
e)399 16 8000

Answer 2

my problem is whether i should have used nCr methods for these..

Answer 3

i assume the first one you mean \(12!\) right?

Answer 4

@satellite73 then again..you know that's the problem huh. hahha. yes!

Answer 5

oops. yeah i didn't mean imaginary numbers lol.

Answer 6

ok now let us dispense with formulas and think only using the counting principle
notice you used no formula for the first one, 12 choices, then 11, then 10, ...

Answer 7

okay... yeah i just did 12 x 11 x ....etc

Answer 8

in the second one the coach is at either and of the bench
how many way if he is in the first seat?

Answer 9

i said two ways...front or back so i just calculated 11! then multiplied that by 2

Answer 10

that is what i would do and notice we appealed to no formula

Answer 11

true....so none of these use a formula at all?? just the counting/multiplication?

Answer 12

unfortunately i do not get the same number you wrote

Answer 13

http://www.wolframalpha.com/input/?i=2*11!

Answer 14

i think maybe you got b and c backwards

Answer 15

oops that is what i got i typed it in differently than i had on my sheet

Answer 16

but i got the same for c

Answer 17

so you were thinking clearly for C as well. treat travis and pat as one person, then multiply by 2

Answer 18

yep... and for D 3 as 1 person ... 6*10!

Answer 19

so answer for b and c are both \(2\times 11!\)

Answer 20

right
and why? because \(3\times 2=6\) possible arrangements for the three of them and then \(10!\) ways to put them in the slots
so once again it is counting principle all the way

Answer 21

Yes that is what I got..cept I did an amateur tree diagram :P

Answer 22

w/e it worked

Answer 23

basically what i DONT get is why this question all uses counting and not nCr formulas.....uuurggh this is frustrating

Answer 24

not sure what that is, but ok
now for the last one i do not know a snap way to do it

Answer 25

what i am trying to suggest is that it will be easier for you (and you seem to have a decent handle on this) if you stop thinking along the lines of "what formula should i use" and keep thinking "what is going on and how to i count this?"

Answer 26

okay I did 11! leaving one or the other out...than assumably theres 12 places to put the other person minus 2 because there are two places they can't be... so I multiplied 11! by 10

Answer 27

you use the formula you need when you need them, but don't get married to any formula. each one is based on the counting principle

Answer 28

right..makes sense

Answer 29

really it is true and in fact one should never even compute the formula as written
suppose i want \(\dbinom{10}{7}\) what do i do? first i think " choosing 7 from 10 is identical to choosing 3 from 10" then i think
10 choices for first slot, 9 for second, 8 for third, all divided by number of ways to arrange these three
then i write
\[\dbinom{10}{7}=\frac{10\times 9\times 8}{3\times 2}=10\times 3\times 4=120\]

Answer 30

thats true. but all too often teachers want you to use formulas :P

Answer 31

math teachers suck in general, but in an on line class who cares
in an on line class i would compute 10 choose 3 like this
http://www.wolframalpha.com/input/?i=10+choose+3

Answer 32

i think we need to be careful with this one
e) if Travis and Pat must be kept separated?

Answer 33

yeah exactly..well i've had to resort to places like this (not in a bad way) because the teacher doesn't answer questions...i understand its summer but i'm a paranoid nerd who needs constant help lol

Answer 34

okay..assumably my answer is wrong

Answer 35

oh i don't know, i didn't do it

Answer 36

i just did 10x11!

Answer 37

but i think the number of possibilities is different depending on whether travis is in the first bench or in the middle

Answer 38

..urggggg

Answer 39

if travis in in the first bench , then in the second bench there are 10 choices and then \(10!\) from there on in

Answer 40

likewise if travis is in the last bench

Answer 41

right, makes sense..

Answer 42

but it travis is in the second bench, then there are 10 choices in the first bench, 9 choices for the third, and then 9! from there on in
so i think you have some work to do

Answer 43

it would appear so.....

Answer 44

but it is not too bad

Answer 45

okay so wait

Answer 46

okay no I don't get it.. never mind -.-

Answer 47

if we label the benches 1 through 12 and put travis in a specific bench, we can compute easily enough
bench 1 or 12 \(10\times 10!\)

Answer 48

10 x 10! makes more sense... i've resorted to drawing out benches which is relatively useless

Answer 49

then in the other boxes i think it is a different calculation

Answer 50

i am sorry i am so confused on this one

Answer 51

box 2 would be 10 choices for bench 1 and then 9 choices for bench 3 and then 9! choices for the remaining 9 benches
does that look rigth?

Answer 52

:(

Answer 53

lets put travis in bench 2
_ T _ _ _ _ _ _ _ _ _ _

Answer 54

okay

Answer 55

how many choice for bench one?

Answer 56

10

Answer 57

that is what i get
how many for bench 3?

Answer 58

also 10

Answer 59

..i think

Answer 60

i think maybe 9

Answer 61

we have travis seated
we have one other person seated
and we cannot seat Pat

Answer 62

couldn't the options for 1 and 3 be interchangeable

Answer 63

yes, but i am just counting as usual
ten choices for first seat, anyone but travis or pat
one choice for second seat, travis
9 choices for third seat not the guy in the first seat, not travis, not pat

Answer 64

oh ok

Answer 65

so thats where you got 10 x 10!

Answer 66

that is where i got \(10\times 10!\) from in the first case , if travis is in seat one

Answer 67

wait but this will be different

T _ _ _ _ _ _ _ _ _ _ _

Answer 68

yes, that will be the \(10\times 10!\) i think
ten choices for bench 2 (everyone but travis or pat) then the remaining ten seats with no restrictions so \(10\times 10!\)

Answer 69

so they both are the same..?

Answer 70

for this one
_ T _ _ _ _ _ _ _ _ _ _
i get
\[10\times 9\times 9!=9\times 10!\]

Answer 71

okay so...... 9 x 10 ! + 10 x 10!

Answer 72

2 min..grilled cheese is calling me

Answer 73

here is my reasoning
bench one
not travis, not pat, ten choices
bench two
travis one choice
bench 3
not travis, not pat, not guy if first bench, 9 choices
other 9 benches, no restrictions, 9! choices

Answer 74

mmm lunch

Answer 75

oh hold one what a much easier way to do this!! now i feel like a moron
take your answer to the first one "a) with no restrictions." and subtract your answer to the third one "c) if Travis and Pat must stand together?"
sorry i did not see this right away
now you are done

Answer 76

seriously :o

Answer 77

i got.... 39916800.
ohdear. that was my original answer :P

Answer 78

thank you sooo much tho!

Answer 79

well i do feel silly for not seeing this right away

Answer 80

i'm going to give up on this assignment now and pray the rest of the questions are right. getting quite sick of it. thank you so much for all your help!

Answer 81

you are quite welcome