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# I got 8 possibilities...it is incorrect.

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+20

How many ways are there to assign each of 6 friends to either the chemistry class or the biology class if one of these six, Manoj, refuses to be in a class without any of his friends?

May 27, 2018

#1
+97576
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How many ways are there to assign each of 6 friends to either the chemistry class or the biology class if one of these six, Manoj, refuses to be in a class without any of his friends?

Yes there are many more than 8 possibilities.

I think I will assign the 5 first and then assign the sixth one later.

number of ways that there can be all 5 in one subject is   1*2 = 2    (because there are 2 subjects)

Manoj would have to be with the others so when he is added there are still only 2 ways

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number of ways that there can be 1 in one subject and 4 in the other in one subject is   5*2 = 10

number of ways that there can be 2 in one subject and 3 in the other in one subject is   5C2*2 = 10*2=20

So that is 30 ways the 5 can be assigned if they are not all in the same class.

with each of these Manoj can be in either class so when I add him the number of possibilities double.

So that is 60 ways where they are not all in the same class.

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Add the other 2 and I think there are   62 ways that the children can be placed into subjects where Manoj is not left on his own.

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If you have been given an answer then please let me know what it is. If it is different from mine maybe we can work out why.

May 27, 2018
#2
+1339
+2

The six (6) friends are assigned to one of two (2) classes. Five (5) are indistinguishable and one (1) is distinguishable. Restrictions: The distinguishable friend must have at least one (1) indistinguishable friend as a classmate.

Remove the distinguishable friend from the set. Distribute the five (5) indistinguishable to the two classes.  Then assign Manoj only to the classes where one (1) or more of his friends are assigned.

$$\left( \binom{c+f-1}{f}\right) \tiny \textbf{ c=number of distinguishable classes & f=number of indistinguishable friends}\\ \left( \binom{2+5-1}{5}\right) * 2 = 12 \tiny \textbf{ Multiply by 2 because each division creates two sets—one for each class. }\\$$

These 12 sets will range from zero to five and five to zero. There are two (2) arrangements where a class has zero (0) of Manoj’s friends, leaving 10 class arrangements for Manoj to join.

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This is also an integer partition of five (5) divided into two sets. Each set represents the two classes with five to zero and zero to five friends.

{5+0}, {4+1}, {3+2}, {2+3}, {1+4}, {0+5}

The resolution is the same as above.

GA

May 28, 2018
#3
+97576
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5 are not indistinguishable Ginger. They are all different people.

Melody  May 28, 2018
#4
+1339
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More advanced combination/partition set questions will introduce sets with elements that are individually distinguishable. This requires the student-mathamation to discern the relevance.

For this question, the five could all be different or they could be genetic clones; it’s irrelevant to the question. For this question, they are Manoj’s friends and are indistinguishable from each other.

Of the six friends, only Manoj is distinguishable from the other five. He is distinguishable for the others because he will only attend a class if (and only if) at least one of his friends also attends.  This is why he is separated (distinguishable) from the others. The other individual characteristics of the set of friends are irrelevant.

Here is a link to a question requiring discernment in what is distinguishable and indistinguishable for the solution set count.  There are two distinguishable sets in this question and the elements of each set are indistinguishable from each other. The correct solution requires the two sets to be treated a one set of indistinguishable elements and partitioned.   (I note you’ve already commented on that post).

GA

GingerAle  May 28, 2018