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 _________.

Let $$a,\,b,\,c$$ denote the numbers of questions chosen from Sections A, B and C respectively.

Total questions to be attempted: $$a+b+c = 15$$.
Minimum requirement: $$a \ge 4,\; b \ge 4,\; c \ge 4$$.
Upper limits (because of the number of questions present): $$a \le 8,\; b \le 6,\; c \le 6$$.

Introduce new variables to handle the “at least 4” condition:
$$a = a' + 4,\; b = b' + 4,\; c = c' + 4,\qquad a',b',c' \ge 0$$.

Substituting in $$a+b+c=15$$ gives
$$a' + b' + c' = 15 - 12 = 3 \qquad -(1)$$

Equation $$(1)$$ demands non-negative integer solutions of sum $$3$$. These are:

• $$\,(3,0,0)$$ and its permutations  • $$\,(2,1,0)$$ and its permutations  • $$\,(1,1,1)$$.

Now apply the upper bounds.

Case 1: $$(a',b',c') = (3,0,0)$$
    Then $$(a,b,c)=(7,4,4)$$ which satisfies $$a\le 8,\;b\le 6,\;c\le 6$$.
    The other two permutations give $$(a,b,c)=(4,7,4)$$ or $$(4,4,7)$$, both violating $$b\le6$$ or $$c\le6$$.
    Hence only one admissible triple: $$(7,4,4)$$.

Case 2: $$(a',b',c') = (2,1,0)$$
    All six permutations obey the limits, giving
    $$(6,5,4),\;(6,4,5),\;(5,6,4),\;(4,6,5),\;(5,4,6),\;(4,5,6).$$

Case 3: $$(a',b',c') = (1,1,1)$$
    This single solution yields $$(5,5,5)$$, within the limits.

Thus the admissible $$(a,b,c)$$ triples are

$$(7,4,4),\;(6,5,4),\;(6,4,5),\;(5,6,4),\;(4,6,5),\;(5,4,6),\;(4,5,6),\;(5,5,5).$$

For each triple, multiply the individual combinations:
  Section A: choose $$a$$ out of $$8 \;\;(\binom{8}{a})$$
  Section B: choose $$b$$ out of $$6 \;\;(\binom{6}{b})$$
  Section C: choose $$c$$ out of $$6 \;\;(\binom{6}{c})$$.

Compute every case:

$$\begin{aligned} (7,4,4):&\;\binom{8}{7}\binom{6}{4}\binom{6}{4}=8\cdot15\cdot15=1800\\[2pt] (6,5,4):&\;\binom{8}{6}\binom{6}{5}\binom{6}{4}=28\cdot6\cdot15=2520\\[2pt] (6,4,5):&\;28\cdot15\cdot6=2520\\[2pt] (5,6,4):&\;\binom{8}{5}\binom{6}{6}\binom{6}{4}=56\cdot1\cdot15=840\\[2pt] (4,6,5):&\;\binom{8}{4}\cdot1\cdot6=70\cdot6=420\\[2pt] (5,4,6):&\;56\cdot15\cdot1=840\\[2pt] (4,5,6):&\;70\cdot6\cdot1=420\\[2pt] (5,5,5):&\;56\cdot6\cdot6=2016 \end{aligned}$$

Add the eight results:
$$1800+2520+2520+840+420+840+420+2016 = 11376.$$

Hence, a student can select the required 15 questions in $$\mathbf{11376}$$ different ways.