Suppose in a unique Soccer series between Earthlings and Martians, the tournament will continue till a team wins 5 matches. Then the number of ways the series can be won by Earthlings, if no match ends in a draw.

Thank you in advance.

My work:

Earthlings can win the tournament in a minimum of 5 matches and a maximum of 9 matches from my logic. I think that we will have to make 5 different cases of 5,6,7,8,9 games considering win in the last game.


  1. 5 matches:- 1 combination $5 \choose5$ =1
  2. 6 matches:- 6th match will be won by earthlings in order to win the series, so we will choose 4 wins from 5 remaining matches i.e. $5 \choose 4$=5
  3. 7 matches:- 7th match will be fixed, so we'll choose 4 from remaining 6 i.e. $6 \choose 4 $=15
  4. 8 matches:- similarly $7 \choose4$=35
  5. 9 matches:- Finally $8 \choose4$=70

Total no. of ways $=1+5+15+35+70=126 $

  • 1
    $\begingroup$ Why downvote ??? It is a legit Question. $\endgroup$ – Dheeraj Kumar Dec 12 '14 at 13:38
  • $\begingroup$ I didn't downvote, but I suppose people want some context- please show your work. $\endgroup$ – voldemort Dec 12 '14 at 13:44
  • 4
    $\begingroup$ It is a legit question, but it is not up to site standards. Questions posted here must contain some work you alredy tried to do before you got stuck. $\endgroup$ – 5xum Dec 12 '14 at 13:44
  • $\begingroup$ Imagine that there are 9 matches. This gives 2^9 = 512 possible win scenarios. Hope this helps. - Matt $\endgroup$ – user145600 Dec 12 '14 at 14:09
  • $\begingroup$ thanks for the help @user145600 but we will also have to consider that team E wins exactly 5 matches $\endgroup$ – Dheeraj Kumar Dec 12 '14 at 14:11

Okay. First we have have to imagine what the possible end-scores can be, if the earthlings have to win. There are of course 5 different end-scores, 5-0,5-1,5-2,5-3 and 5-4. We will now proceed by examining how many ways the different end-scores can occur.

To solve this we will use the binomial coefficient.(http://en.wikipedia.org/wiki/Binomial_coefficient)

$\left( \begin{array}{c} n\\ k\\ \end{array} \right)=\frac{n!}{k!(n-k)!}$

Since the binomial coefficient is exactly the number of distinct ways we can "pick out" k elements of n elements. If you're having trouble imagining this, try to think about it as a line of n white marbles. The binomial coefficient $\left( \begin{array}{c} n\\ k\\ \end{array} \right)$, then gives you the number of ways that you can color k marbles and obtain an uniq line of white and colored marbles.

In your case we the number of combinations where the earthlings wins, is given by the number of ways that they can obtain the end-scores. The number of ways that the earthling can win with end-scores 5-k is given by:

$\left( \begin{array}{c} 5+k\\ k\\ \end{array} \right)=\frac{(5+k)!}{k!(5+k-k)!}$

Because in tournament with end-score 5-k, the number of total matches is 5+k. The number of ways that the martian can win, is equal to the number of ways we can pick out their "k" victories of the total of 5+k matches. Since the earthlings have to win the last match in the tournament, we must have that: "In the case where the martians have won at least 1 game, we will add -1 to the number of total games." In this way we can be sure that the last game is always won by the earthlings.

The total number of combinations, then becomes:

$\left( \begin{array}{c} 5+0\\ 0\\ \end{array} \right)+\left( \begin{array}{c} 5+1-1\\ 1\\ \end{array} \right)+\left( \begin{array}{c} 5+2-1\\ 2\\ \end{array} \right)+\left( \begin{array}{c} 5+3-1\\ 3\\ \end{array} \right)+\left( \begin{array}{c} 5+4-1\\ 4\\ \end{array} \right)$

Which equals:


  • $\begingroup$ You haven't considered E has to win the final match. $\endgroup$ – Rammus Dec 12 '14 at 14:17
  • $\begingroup$ @Rolighed you have not considered that earthlings should also win their last match to win the tournament. $\endgroup$ – Dheeraj Kumar Dec 12 '14 at 14:17
  • $\begingroup$ Oh... you're right :) $\endgroup$ – Rolighed Dec 12 '14 at 14:23
  • $\begingroup$ @Rolighed You answer is absolutely correct!!!!!! i thought of doing it that way but messed up in calculations. $\endgroup$ – Dheeraj Kumar Dec 12 '14 at 14:35
  • $\begingroup$ I think so too :) Glad I could help :) $\endgroup$ – Rolighed Dec 12 '14 at 14:38

Hint: Presumably you are expected to differentiate the number of orders of wins. For example, in a seven game series, EEEMMEE is different from EMEMEEE. How many ways are there for the Earthlings to win in seven games? Remember that the Earthlings have to win the last one.

  • $\begingroup$ yes @RossMillikan i am considering that Earthlings have to win the last one but still no success. $\endgroup$ – Dheeraj Kumar Dec 12 '14 at 14:04
  • $\begingroup$ You will have to add the number of wins for 5,6,7,8,9 game series, treat each case separately and sum them at the end. I would go back and read whatever material you have on combinatorics and make sure you understand how it works if you are struggling. $\endgroup$ – Rammus Dec 12 '14 at 14:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.