Question

Open Challenge:
There are 4 contestants in a race, and ties are allowed. How many different RACE RESULTS are there? Note that it is possible to have a 3-way or 4-way tie.

Answer 1

10?

Answer 2

No, but if you are going to participate in this, you should be prepared to show/present/describe in detail how you got your result.

Answer 3

When you say race result, does that mean 1 contestant winning is different than another contestant winning?

Answer 4

It means to figure out the total number of possible race results. Each race result will be unique. Yes, an A,B,C,D finish is obviously different from an A,C,B,D finish.

Answer 5

Also remember to account for 2-way, 3-way, 4-way ties.

Answer 6

AAAA
AAAB
AABA
ABAA
ABBA
AABB
AABC
ABBB
ACBC
ACCB
ABCC
ACBD
ADBC
ADCB
ACDB
ABDC
ABCD

BAAA
BAAB
BABA
BABB
BABC
BACC
BACD

Im just gonna go ahead and guess cuz my head hurts at the moment, but since the first set that I typed up there^ has 17, multiply that by 4 and you get 68. So is 68 the answer?

Answer 7

A couple of things...
Why did you stop at 17? What is the significance of doing so? Also, if you multiply by 4 you risk overcounting.

Answer 8

Each race result must be unique so I can't agree with multiplying by 4

Answer 9

Also 68 is not the correct result, but nice try. :)

Answer 10

It's 8am here and Im just doin this so I dont do my art work:P

Answer 11

Havent slept yet lol

Answer 12

r there answer choices?

Answer 13

I smell factorials

Answer 14

This is an Open Challenge...no answer choices.

Answer 15

idk maybe 31?

Answer 16

If that were the case, people could just guess the correct answer.

Answer 17

Also @Yuba, if A represents the 1st driver, B represents, the 2nd driver, C represents the 3rd driver and D represents the 4th driver, then AAAA makes no sense.

Answer 18

If you want to represent a 4 way tie, you may have to come up with a more creative way to do so.

Answer 19

AAAA meant they all got first place = 4-way tie lol

Answer 20

Yes, but your approach is not consistent. If that's the case, what would BBBB represent?

Answer 21

A = First, B = Second lol

Answer 22

So there wouldnt have been a BBBB(:

Answer 23

ABCD <<<<these 4 letters can be arranged among themselves in 4! ways.
now tie between 2 people, means there are actually 3 different entities, 2 individual and 1 group of 2, like (AB)CD , and they can be arranged in 3! ways.
similarly, 3-way tie >>>>2! ways
4way tie>>>>1! way = 1 unique way.
in all, 4!+3!+2!+1! ways.

Answer 24

^^^ Won't there be double accounting a few times?

Answer 25

@hartnn, another good try, but it is not the result
@Yuba, I assure you that your approach is limited if you cannot introduce a way to represent each individual racer. You cannot use AAAA to represent 4 different racers.

Answer 26

yeah, not double counting, but i guess, i have missed many cases

Answer 27

like a 3-way tie, (ABC)D, D(ABC)
there are many other cases for this , right?

Answer 28

for a 3 way tie, i think there r 8 cases, 2 with A, 2 with B, 2 with C,2 with D ....not sure i covered everything yet...

Answer 29

you realize this could go on and on and on, right? lol. this could take all night. :)

Answer 30

You can have 1 person win, or 2 persons win,..., or 4. The total number of ways for this is.
$$
\Large{
\Sigma_1^4\binom{4}{n}\binom{3}{3-n}\\
}
$$

Answer 31

@ybarrap, your approach considers "winners" only, but not the total number of race results. There's a huge difference I assure you.

Answer 32

There is also a negative there, impossible. ok

Answer 33

\(\large\binom{4}{n}\) is the number of ways to win, \(\large \binom{3}{4-n}\) is the number of ways for the others to lose:
$$
\Large{
\Sigma_1^4\binom{4}{n}\binom{3}{4-n}\\
}
$$

Answer 34

$$
\Large{
\Sigma_{n=1}^4\binom{4}{n}\binom{3}{4-n}\\
}
$$
Where n = the number of winners, n -=1 means there is 1 winner, n=2 means there is a tie, with 2 winners,..., n=4 means there is a 4-way tie.

Answer 35

You have two groups, winners who may have tied and the rest. The winners could be any of the 4:\(\large \binom{4}{n}\). Once the winners are determined, there are (4-n)! ways that the remaining players can finish. Totals are:
$$
\Large{
\Sigma_{n=1}^4\binom{4}{n}(4-n)!\\
}
$$

Answer 36

@ybarrap, I think you are assuming a specific formula can be applied to this problem to solve it.

Answer 37

I'm actually trying to count the ways that players can place. If you have a tie with 2, you have something like: ww23 or 2w3w. Where "w" means win and 2 and 3 means 2nd and 3rd place.

Answer 38

So \(\binom{4}{1}\) counts things like, w234, 2w34, 23,w4. Those 2nd 3rd and 4the places can be any of 3! positions while wins can only be in 4 positions. So total for a single winner would be 4 \(\times\) 6 = 24.

Answer 39

thats just like arranging 4 people in 4 positions = 4! =24 ways

Answer 40

For ties with 2 winners, you have \(\binom{4}{2}\), with things like: ww23 or w2w3, etc. So there are 6 ways to have ties with 2 winners and 2! ways for the losers to place for a total of 6 \(\times\) 2=12.

Answer 41

@ybarrap , what are you getting for a 3 way tie ?
and for a 2 way tie, there are much more than 12 ways....

Answer 42

For a 3-way ties, I get things like: www2 or w2ww, and 2www. So there are \(\large \binom{4}{3}=4\) ways for the winners and for losers, 1! ways for them to place, for a total of 4 \(\times\) 1=4.

Answer 43

there r 8 possibilities for a 3 way tie,
(ABC)D , ABC coming 1st, D coming 2nd
D(ABC), D coming 1st, ABC coming 2nd.
A(BCD)
(BCD)A
B(ACD)
(ACD)B
C(ABD)
(ABD)C

Answer 44

Hold on people, I'm here.

Answer 45

^like a boss