A LEVEL MATHEMATICS

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Wiped out over tiny P&C problem

Posted by White Group Mathematics on April 4, 2018 at 2:05 AM

The problem is presented as follows:



A student entering a junior college has to offer 4 subjects, of which at least one must be chosen from the basket of science-related subjects ( Maths, Physics, Chemistry And Biology) and at least one being chosen from the basket of arts-related subjects ( Economics, Literature, History And Geography). How many subject combinations in all are available?




Here are my workings.





Case 1



1 art and 3 science subjects : 4 × (4 ×3 ×2)  = 96


(My reasoning: there are 4 possible art subjects to choose from; also each time the student picks one science subject, the number of science subjects still left available for choosing decreases by one, hence the 4 ×3 ×2 portion of the working)



Case 2



2 science and 2 art subjects : (4 ×3 ) ×( 4×3) = 144




Case 3



1 science and 3 art subjects: 4 × (4 ×3 ×2) = 96




Total possible number of combinations = 96 + 144 +96 = 336




However, the actual answer given is merely 68. How is this right? I have scrutinized my workings repeatedly, and can't for the love of god sniff out any mistake. Perhaps the answer supplied is wrong?



Please advise, thanks.





Student X









Hi,


    Your manner of computation unfortunately is erroneous, because it has actually included permutation, ie the sequence in which the subjects are chosen matters based on your workings, when it shouldn't. Please find the amended version below.



Case 1



1 art and 3 science subjects : 4C1  × 4C3 =16



(If I were to permutate accordingly, then I would arrive at 16 × 3! =96, which is equivalent to your answer previously obtained for this particular case)




Case 2



2 science and 2 art subjects : (4C2 ) ×(4C2) = 36



(If I were to permutate accordingly, then I would arrive at 36 × 2!× 2! =144, which is equivalent to your answer previously obtained for this particular case)




Case 3



1 science and 3 art subjects: 4C1 × 4C3 =16



(If I were to permutate accordingly, then I would arrive at 16 × 3! =96, which is equivalent to your answer previously obtained for this particular case)





∴ Total possible number of combinations = 16 + 36 + 16 = 68 (shown)






Hope this clarifies.  Peace.







Best Regards,

Mr Koh



Categories: Math Queries

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