A difficult Perms and Combs Question (1 Viewer)

[hurlstonemaths]

How would you do this question?

"M and F are competing against each other in a competition in which the winner is the first person to score five goals.

The outcome is recorded by listing, in order, the initial of the person who scores each goal.

For example, one possible outcome could be MFFMMFMM."

In how many ways could the outcome of the experiment be recorded?

 

[braintic]

How would you do this question?

"M and F are competing against each other in a competition in which the winner is the first person to score five goals.

The outcome is recorded by listing, in order, the initial of the person who scores each goal.

For example, one possible outcome could be MFFMMFMM."

In how many ways could the outcome of the experiment be recorded?

The minimum number of goals is 5, the maximum is 9.

Assume M wins:
The last goal MUST be M.
There must be EXACTLY 4 Ms before that last goal, out of anywhere between 4 and 8 goals.
4 Ms from 4 goals: 4C4 = 1 way (MMMM)
4 Ms from 5 goals: 5C4 = 5 ways
4 Ms from 6 goals: 6C4 = 15 ways
4 Ms from 7 goals: 7C4 = 35 ways
4 Ms from 8 goals: 8C4 = 70 ways

Add, then double to account for F winning: 252

 

[iStudent]

I'll give this a shot
Let's assume M wins
Then these are the ways the result can be set out:
XMXMXMXMXM
Where X indicates a certain number of "F" positions less than or equal to 4.
There are 4 F's to to be placed in these positions.
Hence, assuming that there are only 1 'f's for each X: (e.g. FMFMFMFMM)
= 5 x 4 x 3 x 2/4! = 5
With 2 'F's' stacked in one of the positions and rest taking up 1 X: (e.g. FFMFMFMMM)
= 5 x 4 x 3/2! = 30
2 F's + 2 F's (e.g. FFMFFMMMM)
= 5 x 4/2 = 10
3 'F's' in one of the positions, other F taking up 1 X (e.g. FFFMFMMMM)
= 5 x 4 = 20
4 F's in one of the Xs (e.g. MFFFFMMMM)
= 5
Total = 70
Since, M can also win, then total = 2x70 = 140

Is that the answer? :s

Oh wait, jokes. I incorrectly assumed F scored 4 times. Yea Braintic probably got it haha

 

[braintic]

I'll give this a shot
Let's assume M wins
Then these are the ways the result can be set out:
XMXMXMXMXM
Where X indicates a certain number of "F" positions less than or equal to 4.
There are 4 F's to to be placed in these positions.
Hence, assuming that there are only 1 'f's for each X: (e.g. FMFMFMFMM)
= 5 x 4 x 3 x 2/4! = 5
With 2 'F's' stacked in one of the positions and rest taking up 1 X: (e.g. FFMFMFMMM)
= 5 x 4 x 3/2! = 30
2 F's + 2 F's (e.g. FFMFFMMMM)
= 5 x 4/2 = 10
3 'F's' in one of the positions, other F taking up 1 X (e.g. FFFMFMMMM)
= 5 x 4 = 20
4 F's in one of the Xs (e.g. MFFFFMMMM)
= 5
Total = 70
Since, M can also win, then total = 2x70 = 140

Is that the answer? :s

Oh wait, jokes. I incorrectly assumed F scored 4 times. Yea Braintic probably got it haha

You say "less than or equal to 4", yet your working assumes EXACTLY 4 Fs.

The 70 that you get is the 8C4 in my solution.

 

[glittergal96]

There is another method:

Each string of 5 M's and 5 F's corresponds to a unique outcome by truncating to the first point where one side has 5 goals (and every outcome has such a 10 character representative string). This is unique because after a player wins, all the remaining characters in the string must be that of the losing player.

So the answer is:

 

[braintic]

There is another method:

Each string of 5 M's and 5 F's corresponds to a unique outcome by truncating to the first point where one side has 5 goals (and every outcome has such a 10 character representative string). This is unique because after a player wins, all the remaining characters in the string must be that of the losing player.

So the answer is:

Now that's the kind of solution I like.

Is there any chance you could come up with something like that for the question posted here:
http://community.boredofstudies.org...20/permutations-combinations.html#post6675645

(Not the original question in the thread, but the one in the 4th most recent post. We've got a solution, but it would be good to get a solution as simple as the one you have here.)

 

[glittergal96]

Now that's the kind of solution I like.

Is there any chance you could come up with something like that for the question posted here:
http://community.boredofstudies.org...20/permutations-combinations.html#post6675645

(Not the original question in the thread, but the one in the 4th most recent post. We've got a solution, but it would be good to get a solution as simple as the one you have here.)

Thanks .

I just tried the question you linked and couldn't think of anything faster than just using the inclusion-exclusion principle. Sorry!

 

[celinelz]

no