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
If all the words with or without meaning made using all the letters of the word "KANPUR" are arranged as in a dictionary, then the word at 440th position in this arrangement is:
The number of 6-letter words, with or without meaning, that can be formed using the letters of the word MATHS such that any letter that appears in the word must appear at least twice, is $ 4 \_\_\_\_\_$.
A | B | C | D | Average |
---|---|---|---|---|
3 | 4 | 4 | ? | 4 |
3 | ? | 5 | ? | 4 |
? | 3 | 3 | ? | 4 |
? | ? | ? | ? | 4.25 |
4 | 4 | 4 | 4.25 |