Discussion
In an examination, six subjects were available for a candidate of which only three had to be offered under the following conditions.
(i) One who offered A had to offer B also and vice-versa.
(ii) One who offered A could not offer E.
(iii) One who offered C or D could not offer F.
The distribution of the candidates over the subjects was as follows.
A - 70; B - 70; C - 90; D - 85; E - 70; F - 35;
How many combinations were permitted?
Answer: A
A takes B and the reverse, A excludes E, and C or D excludes F. The allowed sets of three are:
- with A and B: A B C, A B D, A B F (A B E is barred because A excludes E);
- without A and B, three from C, D, E and F: only C D E, since every other set puts F with C or D.
E appears only in C D E, so all 70 E candidates took C D E. F appears only in A B F, so 35 took A B F. C appears in A B C and C D E, so A B C has 90 − 70 = 20. D appears in A B D and C D E, so A B D has 85 − 70 = 15. Check: A = 20 + 15 + 35 = 70.
So 4 combinations were permitted: ABC, ABD, ABF and CDE. A count of 5 or more comes from allowing A B E, or a set of C, D, E and F that pairs F with C or D, both barred.
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