Question

A student is to answer 10 out of 13 questions is an examination such that he must choose at least 4 from the first five questions. The number of choices available to him is
(a). 196
(b). 280
(c). 346
(d). 140

Answer 1

Hint: In the question, it is given that the student has to choose at least four from the first five questions. This means, the student has two choices – either he can choose 4 out of the first five questions or he can choose 5 out of 5 questions. Based on these choices, we will form cases and then at the end we will add all these cases.

Complete step-by-step answer:

In this question, it is given that the student may either do four out of the five questions or he can do all five in the first five. So depending in this, we will form the two cases:
Case I: A student chooses 4 out of the first five questions. Now if that student chooses 4 out of the first five, he will also have to do some questions from the remaining questions because he has to attend a total of 10 questions in that examination. So the student will have to do six out of the remaining 8 questions. The total number of ways in which he can select 4 out of five questions $$={}^5{C}_4$$ .The total number of ways in which he can select 6 out of the remaining 8 question $$={}^8{C}_6$$.So total number of ways become $$={}^5{C}_4\times {}^8{C}_6$$.
Case II: A student chooses all five out of the first five questions. Now the student must do a total of 10 questions from 13 questions, he will have to do the remaining 5 questions out of the remaining 8 questions. The total number of ways of doing this $$={}^5{C}_5\times {}^8{C}_5$$.
To obtain the total ways in which at least 4 out of the 5 questions can be done, we will add both the cases. Thus we get:
Total ways = Case I + Case II
Total ways $$=\left({}^5{C}_4\times {}^8{C}_6\right)+\left({}^5{C}_5\times {}^8{C}_5\right)$$
Total ways $$=\left(5\times 28\right)+\left(1\times 56\right)$$
Total ways $$=140+56$$
Total ways $$=196.$$
Hence, option (a) is correct.

Note: The above question can also be solved by first finding the total number of ways in which 10 out of 13 questions can be solved and then by subtracting those cases in which we do only 2 and 3 from the first five. Thus we get:
Total ways $$={}^{13}{C}_{10}-{}^5{C}_2\times {}^8{C}_8-{}^5{C}_3\times {}^8{C}_7$$
$$=286-10-80$$
  $$=196$$