Eight people are to be divided into two groups, each having more than

­I have a difference in opinion to DivyaBachu, however, it's mathematics and there can be only one unambiguous answer to a properly worded problem. I did it like this:

8 different people in 2 different groups can be formed in \(2^8 \) ways, but this includes zero people in each group case, so total possible outcomes for us = \(2^8 - 2\)­ = 254

Now, calculating total favorable outcomes is simple. We need 4 people in each group and this can be done in \(^8C_4\) ways­ = \(\frac{8*7*6*5}{4*3*2*1}\) i.e. ­70

Probability = \(\frac{70}{254}\) i.e. \(\frac{35}{127}­\)

Please correct me if I am wrong.­ I did not assume that the groups were identical and I think we should not do that as the question doesn't state it explicitly, right?

Nothing in the question stem suggests that the groups are distinguished by anything other than number, so the 8 can be split into:

1,7
2,6
3,5 and
4,4

One person can be selected:

8 ways. The other 7 become selected automatically.

Two people can be selected:

8C2 = 28 ways, the other 6 become selected automatically.

Three people can be selected:

8C5 = 56 ways, the other 5 become selected automatically.

Four people can be selected:

8C4 = 70 ways. However, because the two groups of 4 are being distinguished by their respective numbers where no distinction exists e.g.:

abcd efgh is the same as efgh abcd

The above 70 should be divided by 2:

35

So the total number of groups is:

8+28+56+35 = 127

And the probability is:

35/127

The correct answer is not among the answer choices.

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