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

I solved this question using case by case approach. If anyone has a shorter approach, please feel free to tag me.

Assuming that Messi will take 9 friends (all 5 footballers): 5C5*5C4 = 5
Assuming that Messi will take 9 friends (but 4 footballers): 5C4*5C5 = 5
Therefore for 9 friends, the number of teams possible = 5+5 = 10

If you calculate this up to 4 friends (since he will take at least 4 footballer friends), the number of possible selections is 191.

How many ways are there for him to invite the NON-football players? There are five of them and each could either be in or out. That means there are 2^5 = 32 ways.

How many ways are there for him to invite the FOOTBALL players? Well, he could either invite all five OR invite four of the five.
There is one way to invite all five. There are five ways to invite four of the five.

So, there are 1*32 ways to invite all five football players plus some combination of non-football players. That's 32. But we need to subtract the 1 way of inviting all 10 since he invites at most 9. So, 31.
And there are 5*32 ways to invite four of the five football players plus some combination of non-football players. That's 160.

31+160 = 191

Answer choice C.

So, the group has 10 friends, and he needs to take at most 9. He can take (0,1,2,3,4,5,6,7,8,9) , but then the question also states that out of 10, there are 5 footballers, and 5 who are not footballers, and he needs to take at least 4 footballers.

So, the groups can be in the following manner:
4 footballers and <= 5 Non footballers (1)
OR
5 footballers and <= 4 Non footballers (2)

Using nCr formulae you will get
For (1)
5c4
5c4 x 5c1
5c4 x 5c2
.
.
5c4 x 5c5
= 160

For (2)
5c5 x 5c4
5c5 x 5c3
.
.
.5c5
= 31

Add total number of ways to get = 191 - Answer

group of 4 footballers + group of 5 Five footballers+ 4 footballers and 1-5 other friends +5 footballers and 1-4 other friends
5+1+25+50+50+25+5+5+10+10+5=191
IMO C

Correct answer: choice C

Minimum number of players: 4
Maximum number of players: 9

for 4 players : total no. of combinations = 5C4 = 5
for 5 players : total no. of combinations = 5C4 * 5C1 + 5C5 = 26
for 6 players : total no. of combinations = 5C4 * 5C2 + 5C5 * 5C1 = 55
for 7 players : total no. of combinations = 5C4 * 5C3 + 5C5 * 5C2 = 60
for 8 players : total no. of combinations = 5C4 * 5C4 + 5C5 * 5C3 = 35
for 9 players : total no. of combinations = 5C4 * 5C5 + 5C5 * 5C4 = 10

therefore 5 + 26+ 55 + 60 + 35 + 10 = 191

There are Two Ways
with 5C5 or 5C4
the Equation will be 5C4(5C0+ 5C1+5C2+ 5C3+5C4+ 5C5) + 5C5(5C0++ 5C1+5C2+ 5C3+5C4)
The answer is 5(1+5+10+10+5+1) +1(1+5+10+10+5)
The Answer is 191

Refer to the picture for solution

Answer is C (191)

Lots of cases to solve.

Have a look at the picture for better understanding.

If the half of his 10 friends are football players and he can invite at least 4 football players, then he can invite 4 or 5 football players to Qatar:
1. He invites 4 football players. There are 5 ways of choosing 4 football players from 5 football players. Then he can invite 0, 1, 2, 3, 4 or 5 friends from 5 remaining (non-football player) friends. This is like finding the number of subsets of a set containing 5 items. So he can choose remaining friends in 2^5 = 32 ways. Using multiplication principle, we get 5 * 32 = 160 ways
2. He invites 5 football players. Since there are only 5 football players, there is only 1 way of choosing 5 football players. Next, he can invite 0, 1, 2, 3 or 4 friends from remaining friends. Since we have 5 remaining friends for at most 4 players, we should find the number of subsets of the set containing 5 items less the number of subsets in which there are 5 items. There is only 1 subset in which there are 5 items. So he can choose remaining friends in 2^5 - 1 = 31 ways. Using the multiplication principle we get 1 * 31 = 31 ways
In total there are 160 + 31 = 191 ways. C)

The number of football friends that will go is either 4 or 5.

Case 1. 4 football friends go
Possibilities: 5C4 x ( 5C0 + 5C1 + 5C2 + 5C3 + 5C4 + 5C5 )
= 5 x ( 1 + 5 + 10 + 10 + 5 + 1 )
= 5 x 32
= 160

Case 2. All 5 football friends go
Possibilities: 5C5 x ( 5C0 + 5C1 + 5C2 + 5C3 + 5C4 ) ( Note that max people can be 9. )
= 1 x ( 1 + 5 + 10 + 10 + 5 )
= 31

=> Total possibilities = 160 + 31 = 191

At most 9 means could be 4 or 5 or 6 or 7 or 8 or 9 people

At least 4 from football impplies 4 from football or five from football

Out of 9 ppl- 5C4 5C5 + 5C5 5C4=10
8 ppl= 35
7 ppl= 60
6 ppl = 55
5 ppl = 26
4 ppl= 5

Thus total= 191
Ans C

avigutman Bunuel pls give a shorter approach.

5 friends are football players, of which atleast 4 are to be invited:
If all 5 football players are invited = 5C5
4 non-football players can be invited as atmost 9 can be invited so the non-football player groups can be represented as 5C0+5C1+5C2+5C3+5C4 (not 5C5 because atmost 9 are allowed) = 1+5+10+10+5 = 31
31*1 = 31

If 4 football players are invited = 5C4
5 non football players can be invited, possible groups = 5C0+5C1+5C2+5C3+5C4+5C5 = 1+5+10+10+5+1 = 32
32*5 = 160

Total of 191 different groups of friends can be formed.

Ans - C

following four scenarios will exist

a) only 4 footballer are invited
b) 4 footballer and 5 non footballer friends are invited
c) 5 footballer and 4 non footballer friends are invited
d) only 5 footballer friends are invited

a) 5c4 = 5
b) 5c4( 5c1 + 5c2+5c3+5c4+5c5) = 155
c) 5c5(5c1 + 5c2+ 5c3+ 5c4+ 5c5) = 30
d) 5c5 = 1

5+ 155+30+1 = 191

hence answer should be C

5C4*(5C0+5C1+5c2+5C3+5C4+5C5) + 5C5 * (5C0+5C1+5c2+5C3+5C4) = 5*32+1*31 = 160+31 = 191

Answer C

We know that at max Messi can invite 9 of his friends, however he wants to invite at least 4 football players. so the number of players that can be invited will lie between 4 to 9, both inclusive

Number of ways Messi can invite 4 players = 5C4 = 5
Number of ways Messi can invite 5 players = 5C4*5C1 + 5C5 = 26
Number of ways Messi can invite 6 players = 5C4*5C2 + 5C5*5C1 = 55
Number of ways Messi can invite 7 players = 5C4*5C3 + 5C5*5C2 = 60
Number of ways Messi can invite 8 players = 5C4*5C4 + 5C5*5C3 = 35
Number of ways Messi can invite 9 players = 5C4*5C5 + 5C5*5C4 = 10

If we sum up the above we obtain a total of 191

IMO C

Messi will invite either 4 football players along with 0 to 5 non-football players or 5 football players along with 0 to 4 non-football players.
If 4 football players invited: 5C4*(5C0+5C1+5C2+5C3+5C4+5C5) = 160 ways
If 5 football players invited: 5C5*(5C0+5C1+5C2+5C3+5C4) = 31 ways
Total 191 ways. Hence, answer C.

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