Permutation Combination Question 6
Question 6 - 2024 (30 Jan Shift 2)
In an examination of Mathematics paper, there are 20 questions of equal marks and the question paper is divided into three sections: A, B and C. A student is required to attempt total 15 questions taking at least 4 questions from each section. If section $A$ has 8 questions, section $B$ has 6 questions and section $C$ has 6 questions, then the total number of ways a student can select 15 questions is
Show Answer
Answer (11376)
Solution
If 4 questions from each section are selected
Remaining 3 questions can be selected either in (1,
$1,1)$ or $(3,0,0)$ or $(2,1,0)$
$\therefore$ Total ways $={ }^{8} \mathrm{c}{5} \cdot{ }^{6} \mathrm{c}{5} \cdot{ }^{6} \mathrm{c}{5}+{ }^{8} \mathrm{c}{6}{ }^{6} \mathrm{c}{5} \cdot{ }^{6} \mathrm{c}{4} \times 2+$
${ }^{8} \mathrm{c}{5} \cdot{ }^{6} \mathrm{c}{6} \cdot{ }^{6} \mathrm{c}{4} \times 2+{ }^{8} \mathrm{c}{4} \cdot{ }^{6} \mathrm{c}{6} \cdot{ }^{6} \mathrm{c}{5} \times 2+{ }^{8} \mathrm{c}{7} \cdot{ }^{6} \mathrm{c}{4} \cdot{ }^{6} \mathrm{c}_{4}$
$=56 \cdot 6 \cdot 6+28 \cdot 6 \cdot 15 \cdot 2+56 \cdot 15 \cdot 2+70 \cdot 6 \cdot 2$
$+8 \cdot 15 \cdot 15$
$=2016+5040+1680+840+1800=11376$
