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Re: The 20 people at a party are divided into n mutually exclusi [#permalink]
Is there any easy way?
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Re: The 20 people at a party are divided into n mutually exclusi [#permalink]
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Brent showed you two approaches.

I do not think there is one more.

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Re: The 20 people at a party are divided into n mutually exclusi [#permalink]
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AE wrote:
Is there any easy way?


It is the same way as brent

Take the number of groups as 2. Then you can have 10,10 split. So 2 is a possible solution.

Take number of groups as 3. You can have 7, 7, 6 split also a possible solution.

Take number 4 You can have 5, 5, 5, 5

Clearly all numbers below 6 are possible.

Now if you look at 6 Split is 3, 3, 3, 3, 4, 4.

Clearly the quanity A is less than quantity B is true and quantity A = quantity B is also true. Hence D
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Re: The 20 people at a party are divided into n mutually exclusi [#permalink]
answer: D
At least number of people in a group is 3, considering the constraint in the question that no group number can be more than 1 different from any other group number, the group members can be just in groups of 3,4 or 3,2. We write the possible combination of 2and3 & 3and4 which equal 20:
3*2 + 2*7 = 20 (9 groups)
3*4 + 2*4 = 20. (8 groups)
3*6 +2 = 20 (7 groups)
(There can’t be odd number of 3, because then we can’t have 20)

3*4 + 4*2. (6 groups)
So n can be 6,7,8 or 9, the answer is D.
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Re: The 20 people at a party are divided into n mutually exclusi [#permalink]
HarveyKlaus wrote:
The 20 people at a party are divided into n mutually exclusive groups in such a way that the number of people in any group does not exceed the number in any other group by more than 1.

Quantity A
Quantity B
The value of n if at least one of the groups consists of 3 people
6


A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.


It doesn't say that the group number is distinct. So we can safely assume it could be repetitive.

One option is 2,3,4,5,6 the sum will be 20. Number of groups 5.

If we have 3,3,3,3,4,4 = 12 + 8 = 20 we have 6 groups.

We could also have it 3,3,3,3,3,3,2 = 3*6 + 2 = 20. 7 groups.

Since we have different answers it is D.
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Re: The 20 people at a party are divided into n mutually exclusi [#permalink]
2
The problem can be summed as two equations:-
4a+3b=20; where a is number of 4 member group & b "3" member group
2a+3b=20; where a is the number of 2 member group and b "3" member group.
given b>=1. so in first case (a,b)= (4,2) ; 4+2=6.
in second case (a,b)= (2,7)=9 (4,4)= 8, (1,6)=7

therefor solution is D.
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Re: The 20 people at a party are divided into n mutually exclusi [#permalink]
1
Since there are 3 people in at least one group, and the prompt also indicates no groups can have beyond 1 more or less than that, we have 2 options:

Option 1: groups of 3 and 4
Option 2: groups of 3 and 2

Let's look at option 1:
The only way I can think to make this add up to 20 people is:
4 groups x 3 people each = 12
2 groups x 4 people each = 8
Total number of groups = 6 groups, which equals our quantity B above.

What about option 2:
One possible solution is:
2 groups x 3 people each = 6
7 groups x 2 people each = 14
Total number of groups = 9 groups, which does not equal our quantity B above.

Since we have already found one arrangement that equals 6 and one that does not, we do not need to keep thinking of all possible arrangements. The answer is D, cannot be determined.
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Re: The 20 people at a party are divided into n mutually exclusi [#permalink]
Hi, why can't the number of people in groups be 2,3,4,5,6? That would make 5 groups and satisfy condtions: atleast 3 in one group as well as not exceeding more than 1 in any other group?
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Re: The 20 people at a party are divided into n mutually exclusi [#permalink]
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The information given by Quantity A says that there is at least one group with 3 people, so all possible combinations are either 2 and 3 or 3 and 4.

(1) For the combination of 2 and 3, find the minimum and maximum values ​​of n. The minimum value is obtained when the combination of 3 is the most, and the maximum value is obtained when the combination of 2 is the most.
-minimum: 3 people can form 6 groups at most, in which case 2 people can form 1 group, n=7
-maximum: 2 people can form 7 groups at most, in which case 3 people can form 2 groups, n=9

(2) For the combination of 3 and 4, find the minimum and maximum values ​​of n. The minimum value is obtained when the combination of 4 is the most, and the maximum value is obtained when the combination of 3 is the most. After enumeration, there are only one combination of 3 and 4: 4 groups with 3 people and 2 groups with 4 people, n=6

D is the answer
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