A student must answer 7 out of 10 questions on an examination. If at l

Hi Bunuel,
I have trouble comprehending why the following approach is flawed:

Step 1: Choose any three of the first five questions - \(5C3\)

With that, we are left with 2 questions from first half and five from second half.

Step 2: Choose any four of the remaining 7 questions on the test - \(7C4\)

This leads to the answer \(5C3 * 7C4\).

What am I missing?

Thanks in advance!

We know that we have 10C7 ways to chose 7 questions from 10 which is 120. The information saying at least 3 questions must be answered out of the 5 first questions will add a constraint hence reducing the number of possibilities by 3C5 which is 10 then 120-10 = 110

Answer C

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The short answer is double counting.

It's best not to first pick
from a group and then include the remainder of that group in another selection because there are no constraints on a given selection being picked again.

For example, if the task were to pick 4 total, then your approach would be

5!/3!2! * 7 =70

In that 70 are 2 options to pick 1 from the remaining two in the initial 5, contributing

5!/3!2! * 2 = 20

But, picking 3 and then 1 from the first 5 is really just picking 4 from 5, and the actual result is

5!/4!1! = 5

So the approach repeats the same 5 four times .

The first 3 picked

abc then d for the 4th pick

could also be

acd and then b

The same selection counted twice.


Posted from my mobile device