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

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Re: GMAT Club World Cup 2022 (DAY 6): 11 players of the Portugal National [#permalink] New post   19 Jul 2022, 03:15
Bunuel wrote:
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. \(\frac{110}{11!}\)

 


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out of 11 players any two players whose position needs to be changed can be selected as 11c2, these two selected players can arrange themselves in 2! ways

Also 11 players can arrange themselves in 11! ways.
hence total probability will be 2(11c2)/11! = 110/11!

option E
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Re: GMAT Club World Cup 2022 (DAY 6): 11 players of the Portugal National [#permalink] New post   19 Jul 2022, 03:17
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Total players = 11
Given condition - no two players of the Portugal National Football Team have the same name

Total no of orders these 11 players can stand in line = 11!
No of orders in which exactly 2 players ate not standing in alphabetical order of their names = no. of ways that by switching exactly two players with each other, the players will be standing in alphabetical order of their name = 11C2 = 55

Required probability = 55/11!

OPTION D
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Re: GMAT Club World Cup 2022 (DAY 6): 11 players of the Portugal National [#permalink] New post   19 Jul 2022, 03:42
1
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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 ?

Total possible arrangements for standing sequence (11 players) = 11!

If only TWO players be switched and they get into correct alphabetical arrangement, it means that balance 9 are in alphabetical arragement. So, only 2 players are NOT in alphabetical arrangement, and they need to be switched.

Possibilities for such 2 persons (favourable outcomes)= 11 C 2 = 55

Hence, Probability for favorable outcomes = (11 C2) / 11!
= 55 / 11!

Option (D) is correct option
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Re: GMAT Club World Cup 2022 (DAY 6): 11 players of the Portugal National [#permalink] New post   19 Jul 2022, 06:28
1
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Bunuel wrote:
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. \(\frac{110}{11!}\)

 


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Switching exactly 2 players will result in all players standing in alphabetical order
We need to find probability of above

P = Desirable outcomes / Total outcomes
Total outcomes = 11! (because there are 11 players and all arrangements equal to 11!)
Now let us find out the desirable outcomes

P1 P2 P3 P4 P5 P6 P7 P8 P9 P10 P11

Now, P1 can interchange with P2, P3 and so on till P11 = 10 possible interchanges
P2 with P1 has already been taken into factor in the previous scenario so 9 possible interchanges for P2
Similarly
P3: 8
P4: 7
P5: 6
P6: 5
P7: 4
P8: 3
P9: 2
P10: 1
P11: 0

So desirable outcomes = 10+9+8+7+6+5+4+3+2+1 = 55

Probability = 55/11!

Answer - D
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Re: GMAT Club World Cup 2022 (DAY 6): 11 players of the Portugal National [#permalink] New post   19 Jul 2022, 07:02
1
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Bunuel wrote:
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. \(\frac{110}{11!}\)



 


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Re: GMAT Club World Cup 2022 (DAY 6): 11 players of the Portugal National [#permalink] New post   19 Jul 2022, 07:10
1
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Bunuel wrote:
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. \(\frac{110}{11!}\)


 


This question was provided by GMAT Club
for the GMAT Club World Cup Competition

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Choosing 2 players who are to switch places from the 11 players we have 11C2 = 110/2 = 55
Since we are taking combinations, we are already eliminating the redundant combinations such as AB and BA as we only have to count one
Now all other Nine players are supposed to be at the same place hence their probability is 1

Overall the combinations of the players are 11!
Hence Probability = 55 x 1 x 1 x 1 x 1 x 1 x 1 x 1 x 1 x 1/11! = 55/11!

IMHO Option D
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Re: GMAT Club World Cup 2022 (DAY 6): 11 players of the Portugal National [#permalink] New post   19 Jul 2022, 07:13
Answer is C

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Re: GMAT Club World Cup 2022 (DAY 6): 11 players of the Portugal National [#permalink] New post   26 Oct 2023, 13:08
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Re: GMAT Club World Cup 2022 (DAY 6): 11 players of the Portugal National [#permalink]
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