Question

A student is to answer 10 out of 13 questions in an examination such that they 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:Consider two cases, where he does 4 questions from the first five questions and does all 5 questions as a second case. Find the number of ways in which he can answer for both the cases.

Complete step by step answer:
Total number of questions that the student has = 13 questions
Out of these 13 questions the student has to answer any 10 questions.
It is said that he will do at least 4 questions from the first 5 questions, which means that he may attend 4 questions from the first 5 questions or he may attend all the 5 questions in the first set.
Let us first consider the case where he answers 4 questions out of the first 5 questions. Thus he has to attend 6 more questions from the next set containing 8 questions.
Thus he attends 6 questions from the second set in $^{8}{{C}_{6}}$ ways.
Thus total ways ${{=}^{5}}{{C}_{4}}{{\times }^{8}}{{C}_{6}}$
It is of the form $^{n}{{C}_{r}}=\dfrac{n!}{(n-r)!r!}$
${{=}^{5}}{{C}_{4}}{{\times }^{8}}{{C}_{6}}=\dfrac{5!}{(5-4)!4!}\times \dfrac{8!}{(8-6)!6!}=\dfrac{5!}{1!4!}\times \dfrac{8!}{2!6!}$
$=\dfrac{5\times 4!}{1!4!}\times \dfrac{8\times 7\times 6!}{2\times 1\times 6!}=\dfrac{5\times 8\times 7}{2}=5\times 4\times 7=140$ways.
Similarly, let us find the case where he answers 5 questions of the first set of questions. Thus he has to attend 5 more questions from the next set containing 8 questions.
He attends 5 questions from 1 set containing 5 questions in $^{5}{{C}_{5}}$ ways.
He attends the rest 5 questions from the second set in $^{8}{{C}_{5}}$ ways.
$\therefore$Total ways = $^{5}{{C}_{5}}{{\times }^{8}}{{C}_{5}}$
$=\dfrac{5!}{(5-5)!5!}\times \dfrac{8!}{(8-5)!5!}=\dfrac{5!}{1\times 5!}\times \dfrac{8\times 6\times 7\times 5!}{3!5!}$
$=\dfrac{1\times 8\times 6\times 7}{3\times 2\times 1}=2\times 4\times 7=56$ways.
Thus the total number of ways in which he can attend 10 questions from 13 questions
= 140 + 56 = 196 ways
The number of choices is 196.
Option A is the correct answer.

Note:We can do it in a table format.

A (5 questions)	B (8 questions)	Total Ways
4	6	$^{5}{{C}_{4}}{{\times }^{8}}{{C}_{6}}$
5	5	$^{5}{{C}_{5}}{{\times }^{8}}{{C}_{5}}$

${{\therefore }^{5}}{{C}_{4}}{{\times }^{8}}{{C}_{6}}{{+}^{5}}{{C}_{5}}{{\times }^{8}}{{C}_{5}}=140+56=196$ ways.