Ways to choose 10 questions from a 13 divided into two sets
A test requires to answer 10 questions, choosing atleast 4 from each of Parts A and B. If there are 6 questions in Part A and 7 in Part B, in how many ways can one choose 10 questions?
One way to solve this would be to consider the different possibilities. One can have
4 from Part A and 6 from Part B
or 5 from Part A and 5 from Part B
or 6 from Part A and 4 from Part B.
So the number is ways is $(^6C_4 \times ^7C_5 )+(^6C_5 \times ^7C_5 )+(^6C_6 \times ^7C_4 ) = 266$
I tried solving this another way but I am getting a different answer. Please explain what is the faulty argument in the following- First choose all the compulsory questions. One has to choose 4 from part A. The number of ways to choose would be $^{6}C_4$. From the second set, the number of ways to choose the compulsory questions would be $^{7}C_4$. This leaves two more questions to be answered which can come from either of the sets. As now there are 2 more question left in A and 3 in B, total number of questions to choose from is 5. So the ways to choose this would be $^5C_2$.
Therefore the total ways to do the selection is $^6C_4 \times ^7C_4 \times ^5C_2 = 5250$. Why is the second method wrong?
PS. I know it is the second one that is wrong because the first one is the textbook solution.
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