SOLUTION: 3. A student is to answer 10 out of 13 questions in an examination such that he must choose at least 4 from the first five questions. The number of choices available to him is:

Algebra ->  Permutations -> SOLUTION: 3. A student is to answer 10 out of 13 questions in an examination such that he must choose at least 4 from the first five questions. The number of choices available to him is:      Log On


   

Question 1027001: 3. A student is to answer 10 out of 13 questions in an examination such that he must choose at least 4 from the first five questions. The number of choices available to him is:
Answer by mathmate(429) About Me  (Show Source):
You can put this solution on YOUR website!

Question:
3. A student is to answer 10 out of 13 questions in an examination such that he must choose at least 4 from the first five questions. The number of choices available to him is:

Solution:
This can be solved using cases.
Case 1: Student answered exactly 4 questions out of the first 5
This can be done in 5C4 ways.
The remaining 6 questions are to be chosen from the remainder, namely 8 questions, in 8C6 ways.
Total = 5C4+8C6 = 5+28 = 33 ways
Case 2: Student answered all 5 question from the first 5
Similarly there are 5C5+8C5=1+56=57 ways.
Total number of ways for both cases = 33+57=90 ways.
Note: nCr is the combination of r objects taken out of n, and is defined as
nCr=n!/(r!(n-r)!)