Tricky counting: 5-on-5 soccer game

Brent@GMATPrepNow wrote:In response to knight247's request at https://www.beatthegmat.com/counting-met ... 87-15.html, I created this tricky counting question.

Please note that I believe that this question is beyond the scope of the GMAT.

In a 5-on-5 soccer game, the team consisting of Al, Bob, Carl, Don and Ed scored a total of 6 goals.
In how many different ways could the 6 goals have been distributed among the 5 players?

A) 30
B) 90
C) 120
D) 150
E) 210

OA: E



Cheers,
Brent

Hi,

The solution for these kind of questions can be obtained using the formula (n+r-1)C(r-1), where n represents the identical items to be distributed among r people including 0. The given problem can be solved using the given formula. So the solution will be (5+6-1)C(5-1) = 10C4 which gives the answer as 210 i.e., option E

Regards,
Chetan

Thanks for that Brent. Appreciate it.

Since I'm still figuring out the problem, I thought I'd use the method I normally use to solve this problem. I've used the 'how many numbers less than 100,000 are there whose sum equals six' logic here.

The goals could be scored in the following manner
0 0 0 0 6 which can be arranged in 5 ways
0 0 0 1 5 which can be arranged in 20 ways
0 0 1 1 4 which can be arranged in 30 ways
0 0 0 2 4 which can be arranged in 20 ways
0 1 1 1 3 which can be arranged in 20 ways
0 0 1 2 3 which can be arranged in 60 ways
0 0 0 3 3 which can be arranged in 10 ways
0 0 2 2 2 which can be arranged in 10 ways
0 1 1 2 2 which can be arranged in 30 ways
1 1 1 2 2 which can be arranged in 5 ways

Sum=210 ways. Hence E

@Chetan...Great stuff with the formula bro. How did you come up with that? or did you get it from some source?

Thanks Brent and Knight247!!!

@Knight247, I have learned the basics of the topic in my 12th standard. Also you can get the formulae and other basics of P&C from the book "Quantitative Aptitude for CAT" by Arun Sharma.

Regards,
Chetan

Hey Guys -- you can refer to the following formulas for "Restricted Permutation"


(a) Number of permutations of ‘n’ things, taken ‘r’ at a time, when a particular thing is to be always included in each arrangement

= r n-1 Pr-1

(b) Number of permutations of ‘n’ things, taken ‘r’ at a time, when a particular thing is fixed: = n-1 Pr-1

(c) Number of permutations of ‘n’ things, taken ‘r’ at a time, when a particular thing is never taken: = n-1 Pr.

(d) Number of permutations of ‘n’ things, taken ‘r’ at a time, when ‘m’ specified things always come together = m! x ( n-m+1) !

(e) Number of permutations of ‘n’ things, taken all at a time, when ‘m’ specified things always come together = n ! - [ m! x (n-m+1)! ]


EDIT-- added a jpeg for better readability

Brent@GMATPrepNow wrote:
In a 5-on-5 soccer game, the team consisting of Al, Bob, Carl, Don and Ed scored a total of 6 goals.
In how many different ways could the 6 goals have been distributed among the 5 players?

A) 30
B) 90
C) 120
D) 150
E) 210

OA: E

Cheers,
Brent

Now, I'll approach in this way, 6 goals will "choose" person in the set (A,B, C,D,E)
For example:

A => B B => C => D => E

It means 1st goal choose A, 2nd don't like A move "=>" to B, 3rd choose B, 4th move "=>" and choose C, 5th move and choose E

Another one is

=> => => => E E E E E E (all of 6 goals don't like the four first player and move "=>" to choose E)

So, we have number of combination of "=>" and "players" are [spoiler]10C4 = 210[/spoiler]

IMO, correct answer is E