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Step-by-Step Solution
Step 1: Understand the Problem
A Mathematics question paper has three sections: A, B, and C with 8, 6, and 6 questions respectively.
The student must answer a total of 15 questions, with at least 4 questions from each section.
Step 2: Set Up Variables and Constraints
Let:
x be the number of questions chosen from Section A ( 4 \le x \le 8 )
y be the number of questions chosen from Section B ( 4 \le y \le 6 )
z be the number of questions chosen from Section C ( 4 \le z \le 6 )
Since the student must attempt exactly 15 questions, we have the equation:
x + y + z = 15.
Step 3: Introduce Shifted Variables
To handle the “at least 4 questions” constraint, define new variables:
x = 4 + a,\quad y = 4 + b,\quad z = 4 + c,
where a,b,c \ge 0 . Substituting these into x + y + z = 15 gives:
(4 + a) + (4 + b) + (4 + c) = 15 \quad \Rightarrow \quad a + b + c = 3.
We also have the upper bounds:
a \le 4 since x \le 8
b \le 2 since y \le 6
c \le 2 since z \le 6
Step 4: Find All Valid (a, b, c) Triplets
We need all nonnegative integer solutions to a + b + c = 3 subject to a \le 4,\, b \le 2,\, c \le 2 .
Listing them out:
a=3,\; b=0,\; c=0
a=2,\; b=1,\; c=0
a=2,\; b=0,\; c=1
a=1,\; b=2,\; c=0
a=1,\; b=1,\; c=1
a=1,\; b=0,\; c=2
a=0,\; b=2,\; c=1
a=0,\; b=1,\; c=2
Each valid triplet (a,b,c) corresponds to (x,y,z) = (4+a,\,4+b,\,4+c) .
Step 5: Compute the Number of Ways for Each Valid Distribution
For each valid (x,y,z) , the number of ways to select the questions is:
\binom{8}{x} \times \binom{6}{y} \times \binom{6}{z}.
We then sum these counts over all valid combinations.
Step 6: Calculate and Sum All Binomial Coefficients
Substituting values for x=4+a, y=4+b, z=4+c into the binomial coefficients and summing:
For a=3, b=0, c=0 \;\rightarrow\; x=7, y=4, z=4 :
\binom{8}{7} \cdot \binom{6}{4} \cdot \binom{6}{4} = 8 \times 15 \times 15 = 1800.
For a=2, b=1, c=0 \;\rightarrow\; x=6, y=5, z=4 :
\binom{8}{6} \cdot \binom{6}{5} \cdot \binom{6}{4} = 28 \times 6 \times 15 = 2520.
For a=2, b=0, c=1 \;\rightarrow\; x=6, y=4, z=5 :
\binom{8}{6} \cdot \binom{6}{4} \cdot \binom{6}{5} = 28 \times 15 \times 6 = 2520.
For a=1, b=2, c=0 \;\rightarrow\; x=5, y=6, z=4 :
\binom{8}{5} \cdot \binom{6}{6} \cdot \binom{6}{4} = 56 \times 1 \times 15 = 840.
For a=1, b=1, c=1 \;\rightarrow\; x=5, y=5, z=5 :
\binom{8}{5} \cdot \binom{6}{5} \cdot \binom{6}{5} = 56 \times 6 \times 6 = 2016.
For a=1, b=0, c=2 \;\rightarrow\; x=5, y=4, z=6 :
\binom{8}{5} \cdot \binom{6}{4} \cdot \binom{6}{6} = 56 \times 15 \times 1 = 840.
For a=0, b=2, c=1 \;\rightarrow\; x=4, y=6, z=5 :
\binom{8}{4} \cdot \binom{6}{6} \cdot \binom{6}{5} = 70 \times 1 \times 6 = 420.
For a=0, b=1, c=2 \;\rightarrow\; x=4, y=5, z=6 :
\binom{8}{4} \cdot \binom{6}{5} \cdot \binom{6}{6} = 70 \times 6 \times 1 = 420.
Now we add all these computed values:
1800 + 2520 + 2520 + 840 + 2016 + 840 + 420 + 420.
Summing these gives
1800 + 2520 + 2520 + 840 + 2016 + 840 + 420 + 420 = 11376.
Step 7: Final Answer
Therefore, the total number of ways a student can select 15 questions such that at least 4 questions are chosen from each section is
\boxed{11376}.
