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

Question

A student must answer 7 out of 10 questions on an examination. If at least 3 of the first 5 questions must be answered, how many possible combinations of 7 questions are available to the student?

A. 50 
B. 80 
C. 110 
D. 220 
E. 360

chetan2u · Accepted Answer

two ways-- 
 1) preferred one 
as there is ONLY one possiblity of choosing 7 out of 10, while choosing less than 3 in first five
choose 2 in first 5 and all 5 from later half..
ways = 5C2 = 10..

total ways = 10C3 = 120
ans = 120-10 = 110
C

2) other method.. 
choosing 3 in first five and 4 in remaining 5= 5C3*5C4 = 10*5=50..
choosing 4 in first five and 3 in remaining 5= 5C4*5C3 = 5*10 =50..
choosing all 5 in first five and remaining 2 out of last 5= 5C5*5C2 = 10

total = 50+50+10=110

HarisinghKhedar · Answer

Thanks Chetan!

I think you made a typo while calculating TOTAL for first method. Cheers!

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VibhasJ · Answer

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!

Fama225 · Answer

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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Answer C

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Regor60 · Answer

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

bumpbot · Answer

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

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