Question:

A local restaurant has 16 vegetarian items and 9 non-vegetarian items in their menu. Some items contain gluten, while the rest are gluten-free.
One evening, Rohit and his friends went to the restaurant. They planned to choose two different vegetarian items and three different non-vegetarian items from the entire menu. Later, Bela and her friends also went to the same restaurant: they planned to choose two different vegetarian items and one non-vegetarian item only from the gluten-free options. The number of item combinations that Rohit and his friends could choose from, given their plan, was 12 times the number of item combinations that Bela and her friends could choose from, given their plan.
How many menu items contain gluten?

Updated On: Jan 2, 2025
  • 1
  • 2
  • 3
  • 4
  • 5
Hide Solution
collegedunia
Verified By Collegedunia

The Correct Option is B

Solution and Explanation

Step 1: Calculate the number of combinations for Rohit and his friends. Rohit and his friends chose 2 vegetarian items from 16 and 3 non-vegetarian items from 9:

$\binom{16}{2} \times \binom{9}{3} = \frac{16 \times 15}{2} \times \frac{9 \times 8 \times 7}{6} = 120 \times 84 = 10,080.$

Step 2: Calculate the number of combinations for Bela and her friends. Let g be the number of items containing gluten. The gluten-free items are:

(16 + 9) − g = 25 − g.

Bela and her friends chose 2 vegetarian items and 1 non-vegetarian item from the gluten-free options:

$\binom{16 - g}{2} \times \binom{9 - g}{1}$

Step 3: Relate the two combinations. It is given that:

10,080 = 12 × $\left[ \binom{16 - g}{2} \times \binom{9 - g}{1} \right]$.

Simplify:

$\binom{16 - g}{2} \times \binom{9 - g}{1} = \frac{10,080}{12} = 840.$

Step 4: Solve for g. Expand the combinations:

$\frac{(16 - g)(15 - g)}{2} \times (9 - g) = 840.$

Simplify:

$\frac{(16 - g)(15 - g)(9 - g)}{2} = 840 \implies (16 - g)(15 - g)(9 - g) = 1,680.$

Testing values of g, we find g = 3 satisfies the equation.

Answer: 3

Was this answer helpful?
1
5

Questions Asked in XAT exam

View More Questions