Question

1. In how many ways can four aces be drawn from a deck of cards? (Order is not important.)

2. If a family has four children, in how many ways could the parents have two boys and two girls?

3. A club consists of 8 men and 14 women. In how many ways can they choose a president, vice president, treasurer, and secretary, along with an advisory committee of five people?

Answer 1

How many cards a deck of cards contain?

Answer 2

52

Answer 3

And there are \(4\) aces in that deck. As order is not important, so it is just selection not arrangement.

Answer 4

So, from \(52\) cards, \(4\) aces can be drawn as:
\[\large ^{52}C_{4}\]

Answer 5

I tried that and got 270725, but it said I was wrong

Answer 6

Wait.

Answer 7

Really? Is this not the answer?

Answer 8

Then I must say I am very poor at this concept. :(

Answer 9

Haha! Yeah I tried a bunch of ways and nothing seems to be right! Could you help me with the other two?

Answer 10

eh, I'm with waterineyes
since the order doesn't matter we choose 4 out of 52

Answer 11

did you do the calculations neatly

Answer 12

yep

Answer 13

@xapproachesinfinity answer is what she is written there.

Answer 14

I cross-checked. :P

Answer 15

Yes I did, I'm going to ask my teacher about it tomorrow because it might just be an error with the homework website maybe :/

Answer 16

Do you have answer for that?

Answer 17

For the first one? no

Answer 18

Are you doing Permutations And Combinations or doing Probability?

Answer 19

Permutations and combinations

Answer 20

I have got a link but it is not clear there also, may be you get some help:
http://openstudy.com/study#/updates/4e925b700b8bfb9f23ec9f68

Answer 21

hmmm
you might try 52C4/52c48 i think we need to take care of the other cards arrangements
52C4 is not arranging 4 cards

Answer 22

I saw that and I tried those answers too and none of them worked! :/

Answer 23

can you try my guess! lol

Answer 24

52C4 and 52C/48 both equal 270725

Answer 25

eh I'm dumb they are supposed to be equal lol

Answer 26

If order is not important, there is only one way to draw 4 aces from a standard pack.

Answer 27

Ha ha ha.. :P

Answer 28

Haha no! I appreciate the help!

Answer 29

2. BGBG
BGGB
BBGG
GGBB
GBGB
GBBG

Answer 30

hehehe, what was i thinking about lol
===========
4C4! only one way

Answer 31

@kropot72 is that right? only one way??

Answer 32

Ah I just realized I did the wrong one for problem 2 I meant, If a family has seven children, in how many ways could the parents have two boys and five girls?

Answer 33

0?

Answer 34

After thinking it is one way

Answer 35

No 4c4=1

Answer 36

Yes it's right! Thank you!!!!

Answer 37

You're welcome :)

Answer 38

Could you all help me with my last two questions?

If a family has seven children, in how many ways could the parents have two boys and five girls?

A club consists of 8 men and 14 women. In how many ways can they choose a president, vice president, treasurer, and secretary, along with an advisory committee of five people?

Answer 39

it should be^_^
after hard thinking i realized it is true! sometimes I'm dumb hehe

Answer 40

2 is already answered look above by @kropot72

Answer 41

Yeah but I put the wrong question by mistake

Answer 42

Can anyone tell how it is \(1\) way only? I am not getting it properly.

Answer 43

eh it is the same principle
bbggggg
bgbgggg
and you continue

Answer 44

Okay, first help her with her problems and then we will discuss. :) I will wait till then.

Answer 45

@waterineyes There are 4C52 ways of choosing 4 cards from a standard pack, however only one of these ways consists of 4 aces.

Answer 46

4P22 = 0
and
5c18= 0
so 0 is the answer?

Answer 47

There isn't a formula I could use for the 7 children one?

Answer 48

No you need to do 22p4 and 18c5

Answer 49

@klovic How did you calculate the value of 4P22?

Answer 50

@xapproachesinfinity I am dumb too. *shakes hand with you*. :P

Answer 51

welcome to dumb club then!^_^

Answer 52

so 1504198080? quite a big number haha

Answer 53

Yeah, I am feeling really nice here.

Answer 54

You all shouldn't feel dumb at all, I am more lost with this than all of you!

Answer 55

what did you get for 22p4? that number

Answer 56

It is the correct answer! Thank you! So there's no formula for the children one?

Answer 57

I got zero @xapproachesinfinity

Answer 58

yes there is

Answer 59

nevermind it is 175560 @xapproachesinfinity

Answer 60

No it is not zero
it should be some huge number

Answer 61

What is the formula for the seven children problem?

Answer 62

I could write it all out but that would just take forever

Answer 63

i guess it is permutaions as well

Answer 64

I'm not sure about it, I usually just write them down hehe

Answer 65

@kropot72

Answer 66

A club consists of 8 men and 14 women. In how many ways can they choose a president, vice president, treasurer, and secretary, along with an advisory committee of five people? There are 22P4 was of choosing a president, vice president, treasurer, and secretary. Having chosen the four officers, there are 18C5 ways of choosing the committee. therefore the total required number of ways is: 22P4 * 18C5.

Answer 67

@klovic
"so 1504198080? quite a big number haha"
You are correct!

Answer 68

If a family has seven children, in how many ways could the parents have two boys and five girls? @kropot72 do you know a formula for this?

Answer 69

There are 7C2 possible arrangements of two boys among the seven children.

Answer 70

THANK YOU SO MUCH ALL OF YOU!!

Answer 71

You're welcome :)