GMAT Club World Cup 2022 (DAY 6): 11 players of the Portugal National

for given 11 players total pair possible which can switch sides and fall in alphabetical order is 5
total P =\(\frac{5}{11!}\)
OPTION C

It's not my strong point, but I guess E

If all 11 people have names beginning with the same alphabet, then the two people can be chosen in 11C2 ways, which is 55 ways.
If 10 out of 11 people have names beginning with the same alphabet, then 10C2 = 45 ways.
Similarly, if we add them all, it comes to 220 ways.

The number of ways in which 11 people can be arranged is 11!, but they can be arranged from both sides.
So, the total number of possible arrangements is 2*11!

Hence, the probability is 220/(2*11!) = 110/11!

Hence, option D is the right answer.

Go with D

I lost in the middle.

Correct answer : Choice E

First the denominator = 11 factorial ways
Now for the numerator = for two postions out of 11 players = we can have 11*10 = 110 ways

hence probability = 110 / 11 factorial ways

No of ways to arrange 11 players is 11!
switching exactly two players with each other, to make the players stand in alphabetical order of their names, there are
10+9+8+7+6+5+4+3+2+1 = 55 ways
therefore desired probability = 55/11!
Answer: D

The 11 players can be arranged in 11! ways.

Hence, except for the two players who are in incorrect positions all other players are in correct positions.

Number of ways to select two players = 11C2 = 55

Number of ways to arrange the players = 1

Required probability = 55/11!


IMO D

Total number of ways 11 players can be arranged in a row=11!
If getting each player standing in a row be standing in alphabetical order of their names by exactly switching 2 players with each other, it means all other players are already standing in their respective positions. Thus we have to find how many ways we can select 2 players out of 11.

Number of ways to select 2 players from 11 players= 11C2
Thus required probability=11C2/11!
=55/11!

[quote="Bunuel"]11 players of the Portugal National Football Team are standing in a row in random order. If no two players of the Portugal National Football Team have the same name, what is the probability that by switching exactly two players with each other, the players will be standing in alphabetical order of their names ?

A. \(\frac{1}{11!}\)

B. \(\frac{2}{11!}\)

C. \(\frac{5}{11!}\)

D. \(\frac{55}{11!}\)

E. \(\frac{110}{11!}\)

Total number of ways by which 11 football players can be arranged- 11!
Now to arrange the players in alphabetical order by switching exactly 2 players= selecting 2 players out of 11 players whose positions are wrong*all other positions are fixes=11C2*1
=11!/(10!*9!)
=11*10/2=55
Thus the probability =No.of ways 2 players can be switched/Number of ways 11 players can be arranged=55/11!

Answer: D

11 players can be arranged in 11! ways.
We need to find the number of cases in which, by switching exactly two players with each other, the players will be standing in alphabetical order of their names.
This is equivalent to finding the number cases in which only two players are not standing in alphabetical order of their names.
A B C D E F G H I J K

Player 'A' can be switched by 10 other players.
Similarly, player 'B' can be switched by 10 other players. But since we already counted 'A' to replace 'B' above, we have 9 other options.
player 'C' will have 8 other options available.
player 'D' will have 7 other options available.
.
.
Lastly, player 'J' will have 1 other unconsidered option left i.e., to swap with player K.

Sum of all possible cases = 10+9+8+7..+1 = 10*11/2 = 55

Hence, probability that by switching exactly two players with each other, the players will be standing in alphabetical order of their names = \(\frac{55}{11!}\)

\(\frac{11C2}{11!}\)=\(\frac{55}{11!}\)

Given: 11 players of the Portugal National Football Team are standing in a row in random order.

Asked: If no two players of the Portugal National Football Team have the same name, what is the probability that by switching exactly two players with each other, the players will be standing in alphabetical order of their names ?

Total number of ways in which 11 players of the Portugal National Football Team can stand in a row in random order = 11!
Number of favorable ways = 11C2 * 2! = 55*2 = 110
Since they can stand in increasing or in decreasing alphabetical order.

The probability that by switching exactly two players with each other, the players will be standing in alphabetical order of their names = \(\frac{110}{11!}\)

IMO E

Answer: D

11 players of the Portugal National Football Team are standing in a row in random order. If no two players of the Portugal National Football Team have the same name, what is the probability that by switching exactly two players with each other, the players will be standing in alphabetical order of their names ?

No. of ways of in which players can be arranged = 11!
No. of ways of players standing in alphabetical order = 1

We need to find the no. of derangements where exactly two players can be switched with each other.
First player can switch with 10 other players to get the required derangement.
Now, second player can switch with 9 other players. (with first player already considered)

In this manner, total no. of derangements will be 10 + 9 + 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 = 55

Hence, the required probability = total no. of derangements / total no. of arrangements.
= 55 / 11!

the correct answer is D
we have eleven players we need to chose between 2 so 11C2 which is 55
the probability is 55/11!

We have to select and interchange 2 players out of 11 players.
This can be done in 11C2 ways=55 ways
Total ways of arranging 11 players = 11!
Therefore, required probability= 55/11!
Answer is D

Imo D

Two players are exchanged. It means that the rest of the 9 players are already arranged.
We have to exchange 2 players. 1st player can be exchanged with another 10 players. = 10 ways
Similarly each other player can be arranged with another 10 players
Total no of ways = 10 * 11
But in each counting there will be a repetition so the Total ways will be = (10*11)/2 = 55

Probability = Favorable / Total
= 55/ 11!