Question

Product or Subtraction Rule (Discrete Math)?

Answer 1

There are 10 people in line, where each person is either male or female. How many different lineups are there, where there are either 5 consecutive mean, or 5 consecutive women?

Answer 2

@Hero

Answer 3

Well, they have to be 5 consecutive men or women in a row, which means 5 men need to be behind each other...So if Im reading the question right, there should be only 2 ways if you're allowed to be Symmetric.

Answer 4

thanks @Yuba how would I represent that? like this?
N1 = 10 people in line
N2 = 2 (either male or female)
N3 = 2 (5 consecutive men or women)
N1 + N2 - N3 = 10 +2 - 2(5) = 2
Is this right?

Answer 5

Im honestly not sure :/ I was already unsure with the question itself honestly

Answer 6

511111
151111
115111
111511
111151
111115

That's six possibilities assuming the 5 represents 5 consecutive men
Now multiply that by 2 to get the possibilities for both men and women.
6 x 2 = 12 possibilities.

Answer 7

So addition property 6 + 6 = 12

Answer 8

ohhhh i thought I would like take it as just
5 5 (men women)
5 5 (women men)
so you have to break the rest into 1 because it could either be male or female, is this right? @Hero

Answer 9

The key word phrase is "either...or"
It wants to know the number of possibilities assuming 5 consecutive men or 5 consecutive women.

If we consider the case where there are 5 consecutive men, then that means the women are not required to be consecutive. That's where using the 1 comes in.

Then we consider the case where there are 5 consecutive women and the men are nto required to be consective.

Then we add to get the total number of possibilities of consecutive men or women.

Answer 10

ahhhh, i get it now. Thank you so much. I wish you're my teacher.

Answer 11

I think Open Study provides a means for better communication that in many ways is more effective than a classroom.

Answer 12

The possible race results are:

1st 1st 1st 1st <4 way tie for 1st>
1st 1st 1st 4th <3 way tie for 1st>
1st 1st 3rd 3rd <2 way tie for 1st><2 way tie for 3rd>
1st 1st 3rd 4th <2 way tie for 1st only>
1st 2nd 2nd 2nd <3 way tie for 2nd>
1st 2nd 2nd 4th <2 way tie for 2nd>
1st 2nd 3rd 3rd <2 way tie for 3rd>
1st 2nd 3rd 4th <no ties>

There is only 1 way to get a four way tie: 4C4 or {A B C D}
There are 4 ways to get a 3 way tie for 1st:
1st 3rd
{ABC} {D}
{ABD} {C}
{ACD} {B}
{BCD} {A}

There are 6 ways to get a 2 way tie for 1st and 2 way tie for 2nd:
1st 4th
{AB} {CD}
{BC} {AD}
{CD} {AB}
{AD} {BC}
{AC} {BD}
{BD} {AC}

There are 12 ways to get a 2 way tie for 1st
There are 4 ways to get a 3 way tie for 2nd
There are 12 ways to get a 2 way tie for 2nd as the only tie
There are 12 ways to get a 2 way tie for 3rd as the only tie
There are 24 ways to get a result with no ties

so the total number of possible results is

1 + 4 + 6 + 12 + 4 + 12 + 12 + 24 = 75

Answer 13

You can study that if you want.

Answer 14

@lynncake

Answer 15

Actually, I think I will re-post this as an Open Challenge. It was a true brain wrecker.

Answer 16

But 75 possibilities is correct.

Answer 17

Nevertheless, I'm going to allow the Open Study community to tackle this one.

Answer 18

so this conclude that the problem is a sum. The question asks just to use product and subtraction rule.

Answer 19

thank you ^_^

Answer 20

Hmmm. Interesting.

Answer 21

It is possible it could be solved that way. The approach I took avoids overcounting

Answer 22

I would rather avoid over-counting. Any time you have to subtract, it means you have over counted.

Answer 23

I have a question @Hero
we could have have 1 1 3 3 without a 2nd?

Answer 24

Yes, that is possible. The reason is the 2nd slot is being held by a first place winner.

Answer 25

If two people finish 1st, the 3rd person automatically gets third place, and NOT second.

Answer 26

you said there's 1 way to get four way tie?
1 1 1 1
could it be
2 2 2 2
3 3 3 3
4 4 4 4

Answer 27

What does 1111 represent?

Answer 28

1st 1st 1st 1st

Answer 29

And I suppose you are assuming it is possible for there to be a 2nd 2nd 2nd 2nd. If that's the case, then who finishes first?

Answer 30

ohhhh right >.< no one

Answer 31

person: a b c d
[1st] [2nd] [3rd] [4th] ways
abcd 1
abc d 4
ab cd 6
a bcd 4
a bc d 4
a b cd 4
a b c d 4
= 27

Answer 32

Nice, try but there are some flaws in this approach. You seem to have left out many possibilities.

Answer 33

If there are no ties how many ways can we put 4 participants in the race result?
is this right? 4^4?

Answer 34

@Hero

Answer 35

4P4 = 24 for no ties.

Answer 36

what's P stands for?

Answer 37

permutation

Answer 38

why is it not 4^4?

Answer 39

Because that would create more possibilities than allotted for that specific scenario.

Answer 40

In a best of (2n-1) series, two teams play until one of the teams accumulates n wins. How many different sequences of results are possible? A best of 7 series? A best of 7 series between A and B where A win the 1st game? @Hero

Answer 41

Another counting problem

Answer 42

I did it like this: 2n-1=7 -> n=4 wins
C(n+r-1, r) -> combination
C(4+7-1, 7) = 120 -> for the first question
is this right? @Hero

Answer 43

For this question, I'm pretty sure you have to count.

Answer 44

I don't understand the series and the game. Does a series have 7 games?

Answer 45

The only thing is, A winning and B losing represents the same results
B winning and A losing represent the same results. So it is either 8 possible or 16 possible for a 7 game series.

Answer 46

You're not going to always be able to use combinations/permutations to solve every counting problem. For some of them you're going to have to actually think.

Answer 47

Basketball has seven game series so I'm familiar with it.

Answer 48

Well, actually, now that I think about it, the question might be a bit more involved than I originally thought.

Answer 49

Yeah, I think there are more possibilities

Answer 50

This is a very involved problem so perhaps using combinations and permutations would be a good approach. If only there were some general formulas for dealing with game series.

Answer 51

Was my approach right though?

Answer 52

Let A = Team A wins
Let B = Team B wins

Then some of the possibilities are

ABABABA
BABABAB

Answer 53

AABBAA is a possibility

Answer 54

yeah

Answer 55

AAAA is also a possibility

Answer 56

Because team A can win four in a row

Answer 57

So this is a very involved problem

Answer 58

I'll have to get back to you later on this.

Answer 59

Looks like your 120 possibilities for 7 game series might be correct.