harsh.champ wrote:In a tournament, there are n teams T1 , T2 ....., T with n > 5. Each team consists of k players, k > 3. The following pairs of teams have one player in common:
T1 & T2 , T2 & T3 ,......, Tn − 1 & Tn , and Tn & T1.
No other pair of teams has any player in common. How many players are participating in the tournament, considering all the n teams together?]

(A)n(k - 1)
(B)k(n - 1)
(C)n(k - 2)
(D)k(k - 2)
(E)(n - 1)(k - 1)

The ans. is A.
Seeking help regarding problem approach!!

The total no. of teams = n
The total no. of players in each team = k
Now without any boundation total no. of players = nk
now total no. of common players = n (T1-T2, T2-T3,...............Tn-T1)

SO TOTAL NO. OF PLAYERS WITH GIVEN CONDITION = nk -n = [spoiler]n(k-1)[/spoiler]

The ans. is A.

harsh.champ wrote:In a tournament, there are n teams T1 , T2 ....., T with n > 5. Each team consists of k players, k > 3. The following pairs of teams have one player in common:
T1 & T2 , T2 & T3 ,......, Tn − 1 & Tn , and Tn & T1.
No other pair of teams has any player in common. How many players are participating in the tournament, considering all the n teams together?]

(A)n(k - 1)
(B)k(n - 1)
(C)n(k - 2)
(D)k(k - 2)
(E)(n - 1)(k - 1)

The ans. is A.
Seeking help regarding problem approach!!

There are n teams and each team has k players
The number of players if there were no sharing = n*k
No of players shared = n

Total no of players with sharing = nk -n = n(k-1)
A

Anyone got answer?

harsh.champ wrote:In a tournament, there are n teams T1 , T2 ....., T with n > 5. Each team consists of k players, k > 3. The following pairs of teams have one player in common:
T1 & T2 , T2 & T3 ,......, Tn − 1 & Tn , and Tn & T1.
No other pair of teams has any player in common. How many players are participating in the tournament, considering all the n teams together?]

(A)n(k - 1)
(B)k(n - 1)
(C)n(k - 2)
(D)k(k - 2)
(E)(n - 1)(k - 1)

Hi!

We have variables in the choices, so we can definitely pick numbers to solve this problem.

Let's pick the smallest numbers we can to keep the question manageable: 6 teams and 4 players per team.

Now let's write out our teams, in accordance with the rules:

1: ABCD
2: DEFG
3: GHIJ
4: JKLM
5: MNOP
6: PQRA

We've gone from letter A to R, so that's 18 different players.

Finally, sub in n=6 and k=4 into the choices:

A) 6(3) = 18
B) 4(5) = 20
C) 6(2) = 12
D) 4(2) = 8
E) 5(3) = 10

Only (A) gives us 18 players... done!

The solution provided by harsh.champ is great and correct but a bit complicated to understand,I want to ask if there is any kind of simple solution anyone can give.
calculus problem solver