Let's just take all sequences of 4 Ws and 3 Ls. If we get rid of all ending Ls, each of these sequences matches up exactly with a sequence of 4 Ws and <= 3 Ls, so we're done. This creates a bijection between all valid orders, and sequences of 4 W's and 3 L's. Hence, there are ${7 \choose 4}=35$ possible valid orders.

Call the two teams A and B. Without loss of generality, we assume that A won. The result can be (in the form of A - B) 4 - 3, 4 - 2, 4 - 1, 4 - 0. Obviously A won in the final match. Consider each case separately.

Case 1: The results were 4 - 3. We have to consider the first 6 matches, in which A won 3 and B won 3. There are 6C3 = 6! / [3! * (6 - 3)!] = 20 possibilities.

Case 2: The results were 4 - 2 We have to consider the first 5 matches, in which A won 3 and B won 2. There are 5C3 = 5! / [3! * (5 - 3)!] = 10 possibilities.

Case 3: The results were 4 - 1 We have to consider the first 4 matches, in which A won 3 and B won 1. There are 4C3 = 4! / [3! * (4 - 3)!] = 4 possibilities.

Case 4: The results were 4 - 0. A won all matches, so there is 1 possibility. Hence in total, there are 20 + 10 + 4 + 1 = 35 possibilities.

There are 4 cases : 4 wins 0 loss, 4 wins 1 loss, 4 wins 2 losses, 4 wins 3 losses.

Let's analyze how many ways 4 wins, n losses can happen. The very last game must be a win. Now, in the remaining 3+n games there are 3 wins and n losses. Thus there are ${n+3 \choose 3}$ this can happen. Therefore the number of ways for this problem is simply ${0+3 \choose 3}+ {1+3 \choose 3}+ {2+3 \choose 3}+ {3+3 \choose 3} = 1 + 4 + 10 + 20 = 35$

It's a combinaation problem.

C(n,r) = n!/r!(n-r)!

In this problem:

n = number of possible matches = 4, 5, 6, 7

r = number of win = 4

for n = 4, C(4,4) = 1

for n = 5, C(5,4) = 5

for n = 6, C(6,4) = 15

for n = 7, C(7,4) = 35

total orders = 1+5+15+35 = 56

for (n+1) matches, it has same path order with n matches as muh as order in n maches.

that is:

For 5 matches, it has same path order with 4 matches as muh as 1 order (WWWWL is same path with WWWW) so there is 1 invalid order.

For 6 matches, it has same path order with 5 matches as muh as 5 orders, so there are 5 invalid order.

For 7 matches, it has same path order with 6 matches as muh as 15 orders, so there are 15 invalid order.

Then, total valid orders is:

total orders - total invalid orders = 56 - (1+5+15) = 35 orders

Consider that all permutations must end with a W since the series ends immediately upon the 4th W, so we can focus on 3 wins and up to 3 losses (since 4 losses would indicate the other team having won 4 times) plus a W at the end. This gives us:

WWWW - 1 permutation

WWWL W - with the L going in any of the first 4 slots - $\frac{4!}{3!1!}$ = 4 permutations

WWWLL W - with L's going in any of the first 5 slots - $\frac{5!}{3!2!}$ = 10 permutations

WWWLLL W - $\frac{6!}{3!}{3!}$ - 20 permutations

Adding up we get 1+4+10+20=35 permutations

The possible number of total games are: 4, 5, 6, and 7.

If only 4 games are played, then they have to be WWWW, so there is only 1 way.

If 5 games were played, the last one should be a W. There are 4C1=4 ways to rearrange the other 4 WWWL.

If 6 games were played, the last one should be a W. There are 5C2=10 ways to rearrange the other 5 WWWLL.

If 7 games were played, the last one should be a W. There are 6C3=20 ways to rearrange the other 6 WWWLLL.

Altogether, there are 1+4+10+20=35 orders.

The last game will always be a win, so there are 4 possible combinations with one loss. With two losses, there are 4 places for the second loss if the first is at the beginning, 3 places if the first is after one win, etc., so that there are 4 + 3 + 2 + 1 possible combinations. If there are three losses, if the first is at the beginning there are 4 + 3 + 2 + 1 possibilites as above. If the first is after one win, there are only 3 + 2 + 1, etc., so that there are (4 + 3 + 2 + 1) + (3 + 2 + 1) + (2 + 1) + (1). There cannot be four losses because then the other team would have won, so the final total is [4] + [4 + 3 + 2 + 1] + [(4 + 3 + 2 + 1) + (3 + 2 + 1) + (2 + 1) + (1)] = 35.

we can split the case into 4 cases case 1. if the team have perfect winners ,WWWW the permutation is 1 case 2. if the team has lost once and played 5 times _ _ _ _ W,in the last is always Win.So the permutation is $\binom{4}{1}$ case 3. if the team has lost twice and played 6 times _ _ _ _ _ W,in the last is always Win.So the permutation is $\binom{5}{2}$ case 4.

if the team has lost twice and played 7 times _ _ _ _ _ _ W,in the last is always Win.So the permutation is $\binom{6}{3}$

so the sum is 35

Firstly, take note that for the winning team, the last game must be a win ( $W$ ). This may seem obvious but it is the crux of my solution below. Secondly, recognise that since there must be exactly $4$ $W$ in total, there will be $3$ $W$ before the last $W$ ---- these $3$ $W$ will have to be permutated as follows:

* Case 1 * Let there be $0$ $L$ :

There will only be $\frac{3!}{3!} = 1$ way to win: $WWWW$

* Case 2 * Let there be $1$ $L$ :

There will be $\frac{4!}{3!} = 4$ ways to arrange $3Ws \text{-- which are not fixed -- and } 1L$ .

* Case 3 * Let there be $2Ls$ :

There will be $\frac{5!}{2! \cdot 3!} = 10$ ways to arrange $3Ws \text{ and } 2Ls$ .

* Case 4 * Let there be $3Ls$ :

There will be $\frac{6!}{3! \cdot 3!}$ ways to arrange $3Ws \text{ and }3Ls$ .

Note that since there can't be $4Ls$ before the last $W$ --- if not the team loses ---, the total number of ways $= 1 + 4+ 10 +20 = \boxed{35}$ .

There are $7 \choose 4$ = 35 ways of arranging 4 Ws and 3 Ls. Any Ls after the final W can be ignored, as the team has already won, but that set of 7 letters still provides that solution uniquely, e.g. WWLWWLL represents WWLWW.

All wins in first 4 - 1 5 th one is a win - no of cases..so on

The games cannot continue indefinitely. Since there are no ties, there can be at most $7$ games, whence one of the teams wins $4$ . Let us consider a particular team for winning orders. Note that the last game is always won by it, since whenever it wins $4$ , it wins the series. Since it may win the series in $4$ , $5$ , $6$ or $7$ games & we don't have choice for last game, the required number is $1 + {4 \choose 1} + {5 \choose 2} + {6 \choose 3} = \boxed{35}$ ways.

Any valid order ends with a $W$ . Before that $W$ , There are $4$ possible cases: $0, 1, 2$ or $3$ losses. Make the combinations in each of those $4$ cases, and get $20 + 10 + 4 + 1 = \boxed {35}$ .

There are $4$ Ws in any winning order, up to $3$ Ls can come before the $4$ th W and there are $4$ slots in which to put any of these $3$ L's:

$\_W\_W\_W\_W$

The number of ways to place up to $3$ indistinguishable items (L's) into $4$ distinguishable locations (the spaces between Ws) can be given by

$\sum\limits_{n=0}^3\dbinom{4+3-1-n}{4-1}=\sum\limits_{n=0}^3\dbinom{6-n}{3}=\dbinom{7}{4}=\boxed{35}$ ways

Maximal aranged $W$ and $L$ are 7 for example $WLWLWLW$

There are $5$ cases for that:

case 1

$WWWW$ there is one arange for a team win the world series

case 2

$LWWWW$ for this case, this is not valid order $(WWWWL)$ because the form same with case 1. So, the valid order are

all of arange $LWWW$ . The number of aranged are $C(4,1=4$

Case 3

$LLWWWW$ for this case, If $L$ after the last $W$ are not valid order because the form same with case 1 or case 2. so,

the valid order are all of arange $LLWWW$ . The number of arange are $C(5,2)=10$

Case 4

$LLLWWWW$ for this case, If $L$ after the last $W$ are not valid order because the form same with case 1, case 2, or

case 3. so, the valid order are all of arange $LLLWWW$ . The number of arange are $C(6,3)=20$

The number of valid order are $1+4+10+20=\fbox{35}$

if the team(losing team)loses 0 games, then we have only one way.If that team loses once then we have 4 ways LWWWW, WLWWW, WWLWW,WWWLW. note WWWWL isn`t a valid order. If the team loses twice , then we have 10 ways because its like arranging 3 Ws and 2 Ls into 5 slots ( if we take six slots... the last one is always a W). Similarly take the case of 3 loses. If we take the case of 4 loses, its not appropriate because the opposite team has already won 4 games. So the total is 1+4+10+20=35.