win the match..!!
India and South Africa play one-day international cricket series until one team wins 4 matches. No match ends in a draw. In how many ways can the series be won?
Try more combinatorics problems.
The answer is 70.
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7 solutions
why only 7 matches?
i don't get the point why when the no. of games played is 5, there are just 4 possible combinations...
I got 5 possible ways..
let I and S denote the victory of India and South Africa respectively.
IIIIS
IIISI
IISII
ISIII
SIIII
Similarly I got 15 and 35 combinations for 6 and 7 games played where India won....
Adding which I got 1+5+15+35
56
the case is same for west indies adding outcomes either way, we get..
2(56)=112
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well the first case IIIIS is same as IIII , since after wining 4 matches, the team will be declared the winner.. thus it will have just 4 possibilities. The same applies for other combinations also!!
I got it wrong, but still... The question asks in how many ways can the series be won, which, at least for me, makes 35 the right answer, because it's the same thing as asking "in how many ways can a team win it?". I think that, for 70 to be the right answer, the question should have been "In how many ways can the series end?" or "How many possible outcomes are there?".
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Yeah, agree. I too first wrote 35 as the answer first and then on the 2nd try wrote 70 as the answer.
The maximum number of matches that can be played is 7. In which there should be 4 wins and 3 loses. So the number of ways in which a team can win the series is the number of permutation of 4 wins and 3 loses.
i.e. 4 ! × 3 ! 7 ! = 3 5
Similarly for the other team. We get 7 0 ways.
Wow! Beautiful solution!
Did the exact same!
If 7 is the maximum number matches, when we calculate permutations for 4W + 3L, aren't we shall be wrongly including impossible combinations such as WWWWLLL ? (This is impossible because the series would not progress beyond the 4th game, because a team has won 4 matches and thereby the series).
My question is that we are calculating only for max games case and not any other case. How can this be logically justified?
Further, given the fact that we still arrive at the same (AND CORRECT) answer, is this a coincidence because of the data or just a 'transformation' of the problem statement ?
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it is actually a very witty solution, counting all combination for 4 W and 3L,we will have the WWWWLLL combination,which is the case when one team wins in 4 games. i love this solution!
Taking I for India and S for South Africa.
Suppose South Africa wins the series, then the last match is always won by South Africa.
| Wins of I | Wins of S | Number of ways | |
| (i) | 0 | 4 | 4 − 1 C 0 = 1 |
| (ii) | 1 | 4 | 5 − 1 C 1 = 3 ! 4 ! = 4 |
| (iii) | 2 | 4 | 6 − 1 C 2 = 2 ! . 3 ! 5 ! = 1 0 |
| (iv) | 3 | 4 | 7 − 1 C 3 = 3 ! . 3 ! 6 ! = 2 0 |
⇒ Total no. of ways = 35
In the same number of ways India can win the series.
⇒ Total no. of ways in which the series can be won = 3 5 + 3 5 = 7 0
4 scenarios:(Wins of India,Losses of SA)= (4,0) , (4,1) , (4, 2) (4, 3) {same also applies for (Wins of SA,Losses of India)}
General rule for this problem: The losing team should not win the last match. Thus when 4 matches are played the losing team should not win any match(no of cases = 1)
.When five matches are played the losing team should not win the fifth match(this implies it must win any one of the first four matches;no of cases = 4C1)
When six matches are played the losing team should not win the sixth match(this implies it must win any two of the first five matches;no of cases = 5C2).
So on and so forth.
Thus the number of ways in which any one team can win = 1 + ( 4C1) + (5C2) + ( 6C3) = 35
Therefore the number of ways in which two teams can win = 2 x 35 = 70.
Case 1: India wins
Arrange that I I I I
To make sure that India wins 4 matches, South Africa must win the game at most 3 times. South Africa's victory also do not exceed the 4th India's victory.
So, we have four boxes which we have to fill at most three balls. Rewrite as:
x 1 + x 2 + x 3 + x 4 ≤ 3 , x n ≥ 0
Notice that there exist a non-negative integer which the equation will equal each other, suppose that p is the integer.
x 1 + x 2 + x 3 + x 4 ≤ 3 ⇒ x 1 + x 2 + x 3 + x 4 + p = 3 , x n ≥ 0 , p ≥ 0
There are ( 3 5 + 3 − 1 ) = ( 3 7 ) = 3 5 ways to arrange South Africa's victory
Case 2: South Africa wins
Similar with the first case, we have 35 ways to arrange India's victory.
Since case 1 and case 2 can't happen at the same time, we have 3 5 + 3 5 = 7 0 ways the series be won
suppose you know already which team is going to win.For example WWWLLW denotes the match outcome.Now by pigeon hole prinsiple,if at most 7 game is played we must have a winner.Now playing matches only ends when the winning team wins the last match.So the only possible cases are to play 4,5,6,7 matches.for which you have to consider 4 w, 3w 1L, 3w 2L, and 3w 3L sequences because last one ends with a winning.hence (1+4!/3! +5!/(3!2!)+6!/(3!3!)=35.But winning team can be selected in 2 ways so altogether in 70 yays
India wins in 4 games: There is 1 way this can happen.
India wins in 5 games : India has to win 3 of the first 4, and then win. There are 4C3 ways this can happen. = 4 ways
India wins in 6 games: India has to win 3 of the first 5, then win. There are 5C3 ways this can happen. = 10 ways
India wins in 7 games: India has to win 3 of the first 6, then win. There are 6C3 ways this can happen. = 20 ways.
Similarly for South Africa to win will have 35 ways.
Total = 70 ways