Question

Ok so i think i have the answers but i think its wrong so please help! Btw this is so be solved using Combination Notation (nCr):
The letters A, B, C,D, E, F,G, H, I, and J are written on slips of paper and
placed in a hat.Two letters are then drawn from the hat.
a. What is the number of possible combinations of two letters?
b. How many combinations consist only of the vowels A, E, or I?
c. What is the probability that the letters chosen would consist only of vowels?
d. What is the probability that the letters chosen would be only B, C,D, or F?

Answer 1

ten letter answer to number 1 is 10*9/2=45

Answer 2

10C2

Answer 3

b there are 3

Answer 4

@satellite73 thats what i got :)

Answer 5

c is 3 out of 45

Answer 6

a. 10*9=90
b. 3*2=6
c. 3*2/10*9=1/15
d. 4*3/10*9=2/15

Answer 7

90 is not right. have to divide by 2

Answer 8

a is not 10*9.

Answer 9

@rnoy i got those for c and d but not for a and b.

Answer 10

d 4 choose 2 is 4*3/2=6

Answer 11

so 6/45

Answer 12

b is 3C2 = 3

Answer 13

for a i got 3,6238,800
b. 3
c. 1/15
d. 2/15

Answer 14

a is not right.

Answer 15

10C2 = 45

Answer 16

@polpak i think it is... i need the total number of outcomes! for A its 9 then B 8 the C 7... i think

Answer 17

No, from the set of 10 letters you are choosing 2. There are only 45 different combinations.

Answer 18

There are only 45 different ways you can pick 2 things from a set of 10. 10C2

Answer 19

thats what i got

Answer 20

ab ac ad ae af ag ah ai aj thats 9 and then for b 8 and c 7 and d 6 and so on

Answer 21

@polpak the question asks not only to choose but, the combination is also asked (eg. difference between [A,B],[B,A] should be considered)

Answer 22

Except you are not removing duplicates.

Answer 23

And even permutations only gives 90 different possibilities.

Answer 24

@polapak i did!

Answer 25

they did

Answer 26

ok well a isnt important! well not too much b c and d are a little more important

Answer 27

9 + 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 = 45 = 10C2

Answer 28

its not plus!!!! its times! clover isnt that right?

Answer 29

thats wrong

Answer 30

b is just 3C2. (picking 2 elements from a group of 3).

Answer 31

kas9987 is right....its times

Answer 32

but isnt is ae ai ei?

Answer 33

No it's not. You were enumerating all the different combinations. If you want to know how many there are you add 1 for each combination.

Answer 34

oh sorrry that equals the same thing!

Answer 35

\[ei \equiv ie\]

Answer 36

That's why it's call 'choose'. If you want to know how many ways you can possibly pick k items from a set of n it's nCk

Answer 37

you mean nCr

Answer 38

no. I mean what I said.

Answer 39

i think so but 3C2 = (3*2)/(2*1) which is 6/2=3

Answer 40

Yes. 3 is the correct answer for b, and 45 is the correct answer for a.

Answer 41

im pretty sure its nCr

Answer 42

And what is r?

Answer 43

no its nCr! We just learned this today! just wondering what type of are you/ have you taken/ passed?

Answer 44

I'm just about finished with this nonsense. I've taken discrete math, linear algebra, multi variable calculus, and differential equations. If you want to call it nCr that's fine, but you need to read what I wrote and realize that what you are calling 'r' (the number of items to pick) I called k and it doesn't matter if I called it argleblargle it would mean the same thing.

Answer 45

nCargleblargle is the number of ways you can possibly pick argleblargle items from a set of n items.

Answer 46

ok! we're sorry!

Answer 47

hi

Answer 48

Ok, so now the question is.. How many items do we have to pick from?

Answer 49

(if we're talking about part a)

Answer 50

10 i think... A-j

Answer 51

Right. And how many items are we picking from that set of 10?

Answer 52

2 but one u pick one u only have 9 to choose from... wait! i think u may be right... I'm really sorry! thank you!

Answer 53

I mean we can list out all the combinations if you want.. it'll take a while, but it's not impossible.

Answer 54

im sorry im just a butt head! i was thinking of it was though i had to pick all of the letters out... I'm really sorry!

Answer 55

First all the ones with a:
ab,ac,ad,ae,af,ag,ah,ai,aj
Then all the ones with b (minus the ones with a):
bc,bd,be,bf,bg,bg,bi,bj
Then all the ones with c (minus any with a or b):
cd,ce,cf,cg,ch,ci,cj
etc.
de,df,dg,dh,di,dj
etc.
ef,eg,eh,ei,ej
etc.
fg,fh,fi,fj
etc.
gh,gi,gj
etc.
hi,hj
etc.
ij
1 + 2 + 3 + 4+ 5+ 6 + 7 + 8 + 9 = 10C2 = 45

Answer 56

For part c, how many vowels are there?

Answer 57

really im so sorry! thank you so much for being patient! and a, e and i so 3!

Answer 58

And so how many different possible ways are there to choose 2 from that set of vowels?

Answer 59

hint: you already answered this question.

Answer 60

hi polpak. did you convince them is was
\[\dbinom{n}{k}\] and not permutations yet?

Answer 61

I think so.

Answer 62

whew!

Answer 63

Kas, there are 3C2 = 3 ways to pick 2 letters such that they are only vowels.

But there are 10C2 = 45 different possible ways you can pick 2 letters from the 10 you have.

So the probability will be \[{3C2\over 10C2} = {3 \over 45} = {1 \over 15}\]

Answer 64

For part d the logic will be similar but you have 4C2 different combinations you can make with the letters listed so you have:

\[{4C2 \over 10C2} = {6 \over 45} = {1 \over 7}\]

Answer 65

umm thank you so much but one thing... 6/45 simplified is 2/15 no 1/7... i just wanted to point that out but thank you so much! :)