Problem Solving

For a player to qualify for the finals of the game competition, he has to win three games - NFS, Chess, Scrabble. 26 players won at least one of the three games. 22 won NFS, 17 won Chess, 19 won Scrabble. What is the difference between the maximum and minimum number of players who could qualify for the finals?

IMO 9

Whats OA and source.
Confirm and Ill post the solution.

hkstrz wrote:For a player to qualify for the finals of the game competition, he has to win three games - NFS, Chess, Scrabble. 26 players won at least one of the three games. 22 won NFS, 17 won Chess, 19 won Scrabble. What is the difference between the maximum and minimum number of players who could qualify for the finals?

T = N + C + S - (NC + NS + SC) - 2(NCS).

The big idea with overlapping group problems is to SUBTRACT THE OVERLAPS.
When we add together everyone in N, everyone in S, and everyone in C:
Those in exactly 2 of the groups (NC + NS + SC) are counted twice, so they need to be subtracted from the total ONCE.
Those in all 3 groups (NCS) are counted 3 times, so they need to be subtracted from the total TWICE.
By subtracting the overlaps, we ensure that no one is overcounted.

In the problem above:
T = 26
N = 22
C = 17
S = 19.
Thus:
26 = 22 + 17 + 19 - (NC + NS + SC) - 2(NCS)
(NC + NS + SC) + 2(NCS) = 32.

MAXIMUM:
To maximize the value of NCS, we must MINIMIZE the value of NC + NS + SC.
If NC + NS + SC = 0, we get:
0 + 2(NCS) = 32
NCS = 16.

MINIMUM:
To MINIMIZE the value of NCS, we must MAXIMIZE the value of NC + NS + SC.
Since N=22, the maximum value of SC = 26-22 = 4.
Since C=17, the maximum value of NS = 26-17 = 9.
Since S=19, the maximum value of NC = 26-19 = 7.
Since the maximum value of NC + NS + SC = 7+9+4 = 20, we get:
20 + 2(NCS) = 32.
NCS = 6.

Thus, the maximum difference = 16-6 = 10.

Mitch,


From a strict formula point of few.

n(A U B U C) = n(A) + n(B) + n(C) - n(A N B) -n(B N C) - n(C N A) + n(A N B N C)

or

T = N + C + S - (NC + NS + SC) - 2(NCS)

The definition of n(A) in the sets formula and the definition of 'N' which you are using differs in way that n(A) includes only A and not any intersections.
And so with n(B), n(C), etc.

Am I right?

eaakbari wrote:Mitch,


From a strict formula point of few.

n(A U B U C) = n(A) + n(B) + n(C) - n(A N B) -n(B N C) - n(C N A) + n(A N B N C)

The definition of n(A) in the sets formula and the definition of 'N' which you are using differs in way that n(A) includes only A and not any intersections.
And so with n(B), n(C), etc.

Am I right?

In the formula above:
n(A) includes EVERY element in A -- including those in A and B, those in A and C, and those in all 3 groups.
n(B) includes EVERY element in B -- including those in A and B, those in B and C, and those in all 3 groups.
n(C) includes EVERY element in C -- including those in A and C, those in B and C, and those in all 3 groups.

n(A N B) includes EVERY element in both A and B -- included those in ALL 3 GROUPS.
n(A N C) includes EVERY element in both A and C -- included those in ALL 3 GROUPS.
n(B N C) includes EVERY element in both B and C -- included those in ALL 3 GROUPS.

Thus, the triple-overlap -- n(A N B N C) -- is included in EVERY term.
The first 3 terms ADD n(A N B N C) to the sum 3 times.
The next 3 terms SUBTRACT n(A N B N C) from the sum 3 times.
The last term ADDS n(A N B N C) to the sum 1 time.
The result is that n(A N B N C) is included in the sum -- correctly -- exactly 1 time.

Because the GMAT typically offers information about those in EXACTLY 2 groups -- a value not represented here -- I prefer the formula used in my initial post.