Question

Can someone please help me with combination?

Answer 1

yea

Answer 2

its numbers 4, 5, 8, 12

Answer 3

:)

Answer 4

4 i think ia B because 8 is bigger than 4

Answer 5

ohh.. ok.. thankyou

Answer 6

5 is 5 because (pack of cards) = 52 / 2=26 then 26/5 = 5.2 so just put 5

Answer 7

ok... this is combination right?

Answer 8

idk what your asking

Answer 9

sorry what i mean is - in solving number 8 you use combination? :D

Answer 10

12 choose 8
12 choose 4

these are combinatorics

Answer 11

yea^

Answer 12

n choose k is just n pick k, divided by k!

Answer 13

ohh.. soo how about no. 8? its permutation??

Answer 14

i meant no. 5. sorry

Answer 15

yes

Answer 16

8 is scary lol .... id have to work it paper and pencil

Answer 17

all red cards, thats half the deck ...

Answer 18

26 cards to pick 5 from

Answer 19

and order doesnt matter so its not a permutation, right?

Answer 20

i think youre referring to combination.

Answer 21

yep. so 26 choose 5

Answer 22

coz permutation is concerned with arrangements??

Answer 23

correct, a perm is a reordering of elements ... since order doesnt matter, we have to divide the permutation by k! assuming we have n pick k permutations

Answer 24

\[^nC_k=\frac{^nP_k}{k!}\]

Answer 25

my answer is like a million+ is this correct??

Answer 26

26.25.24.23.22
--------------
5.4.3.2


26.5.24.23.22
--------------
4.3.2


26.5.6.23.22
--------------
3.2


26.5.6.23.11
--------------
3

26.10.23.11

quite possibly lol

Answer 27

not millions tho

260(253) is less then 75000

Answer 28

ohh.. sorry i forgot that you have to divide it with 5! .. ijust divided it with 5.. haha

Answer 29

:)

Answer 30

its 65,780?

Answer 31

thats what im getting

Answer 32

yeay. haha thanks amistre

Answer 33

yw

Answer 34

btw.. in no. 4 is the answer really B???

Answer 35

which is more: 12 choose 8 or 12 choose 4?

Answer 36

12 choose 4??

Answer 37

ok im confused.. and idrk

Answer 38

compute them. the computation process doesnt change any

Answer 39

ok..

Answer 40

wait just a sec

Answer 41

if you know that the binomial rows are symetric in pascals triangle; then 4 and 8 are the same distance from the ends and are therefore equal without any computations needed

but the computations work regardless of flawed memorys

Answer 42

12.11.10.9.8.7.6.5
-----------------
8.7.6.5.4.3.2


12.11.10.9
----------
4.3.2


12 choose 8 reduces to 12 choose 4

Answer 43

reduces is prolly bad wording .. simplifies?

Answer 44

theyre both 495. :O

Answer 45

yep, so they is equal

Answer 46

hahahaha... and to think the question just pours confusion into my head

Answer 47

amistre can you help me with 3 more nos.??

Answer 48

maybe, if 8 is one of them then i aint got a good approach yet.

Answer 49

its no. 7, 8, and 12 .. if its ok with you?

Answer 50

its ok.. :)

Answer 51

number 8 is tricky to me. 9 choose 3 maybe it but im having trouble 'believeing' that

Answer 52

pick any 3 points and they form a triangle, but the order we choose the same set of points doesnt matter so im believeing it more

Answer 53

hahaha.. well i could just skip that.. but how about nos. 7 and 12

Answer 54

7 is simple enough, its just multiplication of choices

Answer 55

choose 2 boys, that leaves 3 girls

choose 2 girls, that leaves 3 boys

Answer 56

i kinda know this but.. im just confused on how to put it on place

Answer 57

\[\binom{b}{2}\binom{g}{3}\]
\[\binom{g}{2}\binom{b}{3}\]

Answer 58

there are 6 boys in total and 3 girls in total ....

Answer 59

ok.. wait i think i got it!

Answer 60

just give me 2 minutes haha

Answer 61

spose there are 4 ways to pick a set

1 2 3 4

and 2 ways to pick each subset

1 2 3 4
1 2 1 2 1 2 1 2

then we have 4 ways to pick 2 ... 2+2+2+2 = 4(2)

Answer 62

so, there are 6 choose 2 ways to pick only 2 boys

and for each set, there is 3 ways to pick 3 girls

its just multiplication then :)

Answer 63

seriously??? :O :O :O

Answer 64

then that means i dont get it.

Answer 65

what is 6 choose 2?
what is 3 choose 3?

Answer 66

i thout the answer in a.) is 35
coz.. what i did is use combination in boys and girls and add them up

Answer 67

adding is bad, since ive been mentioning multiplication

Answer 68

ohhh.. -_- ... im soo dumb. lool

Answer 69

6.5
--- = 6 choose 2 = 15
2

there are 15 ways to choose 2 boys from 6

now for each choice, there are 3 choose 3 girls that can be chosen .... well, 1 way to choose 3 girls from 3 girls sooo ... 15 ways to choose 1 is just

15(1)

Answer 70

wait!!1.. im quite right.. with the boys section but what i did in the girls that led me to "35" is that i use "6 choose 3" - which should be 3 choose 3

Answer 71

-_-

Answer 72

sooo the answer should be 15... :))

Answer 73

choosing 2 girls from 3 is: 3 choose 2 = 3 ways

now there are 6 boys and 3 to choose to fill the spots, 6 choose 3 is:

6.5.4
---- = 20
3.2

3 ways to gather 20 is 3(20)

Answer 74

ohh.. ok.. i got it. :D

Answer 75

num12 tho, im not sure if your material has a thrm, or if just brute mathing it would be acceptable

Answer 76

i think brute.. haha but our teacher says we can use other stuffs to prove it.

Answer 77

i think what he ment was like the pascal's triangle i guess

Answer 78

and we already answered letter a.

Answer 79

B is our homework

Answer 80

\[(a+b)^n =\binom{n}{0}a^nb^0+\binom{n}{1}a^{n-1}b^1+\binom{n}{2}a^{n-2}b^2+...+\binom{n}{n}a^0b^n\]

if a=b=1 then we have \[(1+1)^n =\binom{n}{0}+\binom{n}{1}+\binom{n}{2}+...+\binom{n}{n}\]
\[2^n =\binom{n}{0}+\binom{n}{1}+\binom{n}{2}+...+\binom{n}{n}\]

Answer 81

ohh WOOOWWW :O :O

Answer 82

but the question is slightly modified from this

Answer 83

we have:
\[2^k=\binom{k+1}{0}+\binom{k+1}{1}+...+\binom{k+1}{r}\]
where r might be k/2
this would be what a thrm would present to us, otherwise its just brute mathing it into submission

Answer 84

OHHH...

Answer 85

\[\binom{n}{x}+\binom{n}{x+1}=\binom{n+1}{x+1}\]

1 4 6 4 1
1 5 10 10 5 1

\[\binom{4}{0}+\binom{4}{1}=\binom{5}{1}\]

Answer 86

amistre.. is the process really long?

Answer 87

\[2^k=\binom{k+1}{0}+\binom{k+1}{1}+...+\binom{k+1}{r}\]
\[2^k=\left[\binom{k+1}{0}\right]+\left[\binom{k+1}{1}\right]+...+\left[\binom{k+1}{r}\right]\]
\[2^k=\left[\binom{k}{0}\right]+\left[\binom{k}{0}+\binom{k}{1}\right]+\left[\binom{k}{1}+\binom{k}{2}\right]+...+\left[\binom{k}{r-1}+\binom{k}{r}\right]\]
\[2^k=2\left[\binom{k}{0}\right]+2\left[\binom{k}{1}\right]+2\left[\binom{k}{2}\right]+...+2\left[\binom{k}{r-1}\right]+\left[\binom{k}{r}\right]\]

not too sure how to procede from that, i have an idea, ut nothing fleshed out yet

Answer 88

the typing it out is arduous :)

Answer 89

yeah, im at a loss there for any generality. unless youve got a thrm in your material already stated, brute mathing it out is the only 'simple' way i can see it

Answer 90

hahahha.. but seriuosly though THANKYOU very very much!

Answer 91

*seriously

Answer 92

maybe:

\[2^k = \binom{k}{0}+\binom{k}{1}+\binom{k}{2}+....+\binom{k}{k}\]
\[2^{k+1} = 2\binom{k}{0}+2\binom{k}{1}+2\binom{k}{2}+....+2\binom{k}{k}\]
let k = n+1,
\[2^{n+2} = 2\binom{n+1}{0}+2\binom{n+1}{1}+2\binom{n+1}{2}+....+2\binom{n+1}{n+1}\]

nah, i cant seem to develop a generality from it :)

Answer 93

oh well, just compute it term for term, add them up and compare to the power of 2

Answer 94

\[2^4 = \binom{4}{0}+\binom{4}{1}+\binom{4}{2}+\binom{4}{3}+\binom{4}{4}\]
\[2^4 = \binom{5}{1}+\binom{5}{3}+\binom{5}{5}\]
lol, still not pulling it together