GMAT Club World Cup 2022 (DAY 4): Messi plans a trip to Qatar and

The correct answer is C
At most 9 friends, means 1 friend or 2 friends, ... 9 friends,
But since Messi is taking with him at least 4 footballers then the number of teams of friends he is taking with him is restricted to : 4 friends or 5 or 6 or 7 or 8 or 9

4 friends (at least 4 footballers) = Comb(4 of 5) = 5
5 friends (at least 4 footballers) = Comb(4 of 5) x Comb(1 of 5) +Comb(5 of 5)=26
6 friends (at least 4 footballers) = Comb(4 of 5) x Comb(2 of 5) +Comb(5 of 5)xComb(1 of 5)=55
7 friends (at least 4 footballers) = Comb(4 of 5) x Comb(3 of 5) +Comb(5 of 5)xComb(2 of 5)=60
8 friends (at least 4 footballers) = Comb(4 of 5) x Comb(4 of 5) +Comb(5 of 5)xComb(3 of 5)=35
9 friends (at least 4 footballers) = Comb(4 of 5) x Comb(5 of 5) +Comb(5 of 5)xComb(4 of 5)=10

Total = 5+26+55+60+35+10=191, the correct answer is C

We have two groups in total, 5 Footballers and 5 Non-Footballers.

F F F F F | N N N N N

We know that Messi wants at least 4 Footballers and at most 9 friends. As you notice, we have a lower-bound with minimum of 4 footballers and upper bound of total 9.

Lets split into 2 cases -

Case 1: We select the minimum of 4 footballers and we can select 0 to 5 Non-footballers (As max is 9). We can do it as follows -

5C4 * (5C0 + 5C1 + 5C2 + 5C3 + 5C4 + 5C5) = 5 * (1 + 5 + 10 + 10 + 5 + 1) = 5 * 32 = 160

Case 2: We select all 5 footballers and we can select 0 to 4 Non-Footballers (Remember, 9 is the max. If we select 5 Fs, we can only select a max up to 4 Ns). We can do it as follows -

5C5 * (5C0 + 5C1 + 5C2 + 5C3 + 5C4) = 1 * (1 + 5 + 10 + 10 + 5) = 1 * 31 = 31

Total Cases = Case 1 + Case 2 = 160 + 31 = 191 (Answer C)

Total Footballers (f) among friends = 5
Total Non-Footballers (n) among friends= 5

Possible combinations with given conditions are below in a tabular format:

Total
5 5
F(f) NF(n) Count (5Cn + 5Cf)
---------------------------------------------------
4 0 | 5
4 1 | 25
4 2 | 50
4 3 | 50
4 4 | 25
4 5 | 5
5 0 | 1
5 1 | 5
5 2 | 10
5 3 | 10
5 4 | 5
------------------------------------------------------
TOTAL: 191

Ans C

[quote="Bunuel"]Messi plans a trip to Qatar World Cup 2022 and intends to invite at most 9 of his 10 friends with him. If half of his 10 friends are football players and Messi wants to invite at least 4 football players to Qatar, then how many different groups of friends can Messi form for the trip ?


A. 63
B. 64
C. 191
D. 192
E. 1023


Given statement - the group can be formed with at most 9 friends and at least 4 of them should be football players
Below are the possible ways through which the team can be formed.
1.Group with 4 football players - 5C4=5
2.Group with all 5 football player - 5C5=1
3.Group with 4 football players and 1 friend who is not a player - 5C4*5C1=5*5=25
4.Group with 4 football players and 2 friend who is not a player - 5C4*5C2=5*10=50
5.Group with 5 football players and 1 friend who is not a player - 5C5*5C1=1*5=5
6.Group with 4 football players and 3 friend who is not a player - 5C4*5C3=5*10=50
7.Group with 5 football players and 2 friend who is not a player - 5C5*5C2=1*10=10
8.Group with 4 football players and 4 friend who is not a player - 5C4*5C4=5*5=25
9.Group with 5 football players and 3 friend who is not a player - 5C5*5C3=1*10=10
10.Group with 4 football players and 5 friend who is not a player - 5C4*5C5=5*1=5
11.Group with 5 football players and 4 friend who is not a player - 5C5*5C4=1*5=5
Thus total Number of groups =5+1+25+50+5+50+10+25+10+5+5
=191
C is the answer

IMO Option C 191 is correct.

Let FF = Football Friends and and OT = Other Friends

4 Football Friends =5
4 Football Friends and 1 other Friend 5*5=25
4 Football Friends and 2 other Friends 5*10=50
4 Football Friends and 3 other Friends 5*10=50
4 Football Friends and 4 other Friends 5*5=25
4 Football Friends and 5 other Friends =5

5 Football Friends =1
5 Football Friends and 1 other Friend =5
5 Football Friends and 2 other Friends =10
5 Football Friends and 3 other Friends =10
5 Football Friends and 4 other Friends =5


Add them all together to give 191.

Answer: C

Messi plans a trip to Qatar World Cup 2022 and intends to invite at most 9 of his 10 friends with him. If half of his 10 friends are football players and Messi wants to invite at least 4 football players to Qatar, then how many different groups of friends can Messi form for the trip ?

Messi has 5 FP and 5 NFP as friends.
Maximum 9 friends can be invited out of which minimum 4 needs to FP.

Case 1: when only 4 FP are selected and rest selected are NFP.
4 FP can be selected in 5 different ways.
these 4 FP selection can be combined with minimum 0 NFP and maximum 5 NFP
i) 0 NFP in 5*1 = 5 ways.
ii) 1 NFP in 5*(5C1) = 5*5 = 25 ways.
iii) 2 NFP in 5*(5C2) = 5*10 = 50 ways.
iv) 3 NFP in 5*(5C3) = 5*10 = 50 ways.
v) 4 NFP in 5*(5C4) = 5*5 = 25 ways.
vi) 5 NFP in 5*(5C5) = 5*1 = 5 ways.

Total (Case 1)= 5 + 25+ 50 + 50 + 25 + 5 = 160

Case 2: when 5 FP are selected and rest selected are NFP
5 FP can be selected in only 1 way.
This selection can be combined with minimum 0 NFP and maximum 4 NFP
i) 0 NFP in 1*1 = 1 way
ii) 1 NFP in 1*(5C1) = 1*5 = 5 ways.
iii) 2 NFP in 1*(5C2) = 1*10 = 10 ways.
iv) 3 NFP in 1*(5C3) = 1*10 = 10 ways.
v) 4 NFP in 1*(5C4) = 1*5 = 5 ways.

Total (Case 2)= 1 + 5+ 10 + 10 + 5 = 31

So, the different groups of friends that Messi can form for the trip
= Total (Case 1) + Total (Case 2)
= 160 + 31
= 191

Total no.of different groups = 5c4x(5c0+ 5c1+...+ 5c5)+5c5x(5c0+5c1+....+5c4)
= 5x32+ 1x31
= 191

Posted from my mobile device

5 footballers, 5 Non footballers


Total friends 4 5 5 6 6 7 7 8 8 9 9
Non Footballer 4 4 5 4 5 4 5 4 5 4 5
Footballer 0 1 0 2 1 3 2 4 3 5 4
Total No of ways 5 25 1 50 5 50 10 25 10 5 5 = 191

Ans C

Friends who are football players = 5. Out of these 5, at least 4 must be invited. In total he can invite at most 9 people, which means:

Case 1: 4 football friends are invited. Of the remaining 5 friends, all combinations from none to all are possible.

Total ways possible = 5C4 times [5C0 + 5C1 + 5C2 + 5C3 + 5C4 + 5C5] = 5 times 2^5 = 5*32 = 160


Case 2: 5 football friends are invited. Of the remaining 5 friends, all combinations from none of groups of 4 are possible.

Total ways possible = 5C5 times [5C0 + 5C1 + 5C2 + 5C3 + 5C4] = 1 times [2^5 - 1] = 31

Hence, in total 160 + 31 = 191 different groups of friends are possible. Option (C) is our correct answer.

Posted from my mobile device

My answer is C

Posted from my mobile device

Total friends 10
Football friends - 5
Non-Football friends - 5

So Messi wants to take at least 4 of his 5 football friends and at most a total of 9 friends

Let's take the first option where Messi takes 4 of his 5 football friends then we have the following scenarios
Messi takes 0 Non-Football friends = 5C4 * 1
OR
Messi takes 1 Non-Football friends = 5C4 * 5C1
OR
Messi takes 2 Non-Football friends = 5C4 * 5C2
OR
Messi takes 3 Non-Football friends = 5C4 * 5C3
OR
Messi takes 4 Non-Football friends = 5C4 * 5C4
OR
Messi takes 5 Non-Football friends = 5C4 * 5C5
Total combinations are = = 5C4 + 5C4 * 5C1 + 5C4 * 5C2 + 5C4 * 5C3 + 5C4 * 5C4 + 5C4 * 5C5 = 5C4(1 + 5 + 10 + 10 + 5 + 1) = 160

Similarly, let's take the option where Messi takes all 5 of his 5 football friends. Then we have the following scenarios
Total combinations are = = 5C5 + 5C5 * 5C1 + 5C5 * 5C2 + 5C5 * 5C3 + 5C5 * 5C4 = 5C5(1 + 5 + 10 + 10 + 5) = 31

Total combinations = 160 + 31 = 191

IMHO Option C

Let's do it case by case :

Case 1: Messi inviting 4 people[this has to be all footballers as per condition]

(1) 5C4 = 5 WAYS

Case 2: Messi inviting 5 people

(1) 5C4 X 5C1 = 25 WAYS
(2) 5C5 = 1 WAY (ALL 5 FOOTBALLERS INVIED)

Case 3: Messi inviting 6 people

(1) 5C4 X 5C2 = 50 WAYS
(2) 5C5 X 5C1 = 5 WAYS (ALL 5 FOOTBALLERS INVIED)

Case 4: Messi inviting 7 people

(1) 5C4 X 5C3 = 50 WAYS
(2) 5C5 X 5C2 = 10 WAYS (ALL 5 FOOTBALLERS INVIED)

Case 5: Messi inviting 8 people

(1) 5C4 X 5C4 = 25 WAYS
(2) 5C5 X 5C3 = 10 WAYS (ALL 5 FOOTBALLERS INVIED)

Case 6: Messi inviting 9 people

(1) 5C4 X 5C5 = 5 WAYS
(2) 5C5 X 5C4 = 5 WAYS (ALL 5 FOOTBALLERS INVIED)

ADDING ALL THE POSSIBLE WAYS OF INVITATION = 191 WAYS

IMO Option C

Messi plans a trip to Qatar World Cup 2022 and intends to invite at most 9 of his 10 friends with him. If half of his 10 friends are football players and Messi wants to invite at least 4 football players to Qatar, then how many different groups of friends can Messi form for the trip ?


A. 63
B. 64
C. 191
D. 192
E. 1023

There are 10/2 = 5 Football players; and balance 5 players play other games.

Hence, 2 cases are possible ie,

Case - 1 : 4 football players + Other Games' players (0 / 1 / 2 / 3 / 4 / 5)

Number of possible combinations = 5C4 * (5C0 + 5C1 + 5C2 + 5C3 + 5C4 + 5C5)
= 5 * (1 + 5 + 10 + 10 + 5 + 1)
= 5 * 32
= 160

Case - 2 : 5 football players + Other Games' players (0 / 1 / 2 / 3 / 4)

Number of possible combinations = 5C5 * (5C0 + 5C1 + 5C2 + 5C3 + 5C4)
= 1 * (1 + 5 + 10 + 10 + 5)
= 1 * 31
= 31

Thus,TOTAL possible combinations = 160 + 31
= 191

(C) is the CORRECT answer

Total friends = 10
Total football friends = 5

Minimum invites = 4 (question says Messi wants to invite at least 4 football friends)

Maximum invites = 9 (question says at most 9 friends out of 10)

So at least 4 (out of 5) football friends and at most 9 out of 10

Case 1: 4 football friends and (either 0 or 1 or 2 or 3 or 4 or 5 non football friends)

5c4 * (1 + 5c1 + 5c2 + 5c3 + 5c4 + 5c5) = 5*(32) = 160

Case 2: 5 football friends and (either 0 or 1 or 2 or 3 or 4 non football friends)

5c5 * (1 + 5c1 + 5c2 + 5c3 + 5c4) = 1*(31) = 31

Total = Case 1 + Case 2 = 160+31 = 191

Answer - C

Posted from my mobile device

Footballer - friends who can play football - Total number = 5
Other - friends who cannot play football - total number = 5

He can form groups in 6 following manners :-

4 membered = 4 footballer = 5c4 ways = 5 ways

5 membered = 4 footballer + 1 other OR 5 footballer = [5c4*5c1] + [5c5] = 26 ways

6 membered = 4 footballer + 2 other OR 5 footballer + 1 other = [5c4*5c2] + [5c5*5c1] = 50 + 5 = 55 ways

7 membered = 4 footballer + 3 other OR 5 footballer + 2 other = [5c4*5c3] + [5c5*5c2] = 50 + 10 = 60 ways

8 membered = 4 footballer + 4 other OR 5 footballer + 3 other = [5c4*5c4] + [5c5*5c3] = 25 + 10 = 35 ways

9 membered = 4 footballer + 5 other OR 5 footballer + 4 other = [5c4*5c5] + [5c5*5c4] = 5 + 5 = 10 ways

Total number of ways = 5 + 26 + 55 + 60 + 35 + 10 = 191 ways

IMO The correct answer is D

Imo C

There is a pool of 5 football players and 5 non football players-

Two cases -
1. When Messy Invites 4 football players then
Total no of ways = 5C4* (5C0+5C1+5C2+5C3+5C4+5C5)

2. When Messy Invites 5 football players then
Total no of ways = 5C5* (5C0+5C1+5C2+5C3+5C4)

Overall: 191

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