Key Concepts and Formulas

• Combinations: The number of ways to choose $r$ objects from a set of $n$ distinct objects without regard to order is given by ${}^nC_r = \frac{n!}{r!(n-r)!}$.
• Multiplication Principle: If there are $m$ ways to do one thing and $n$ ways to do another, then there are $m \times n$ ways to do both.
• Addition Principle: If there are $m$ ways to do one thing and $n$ ways to do another, and the two things cannot be done at the same time, then there are $m + n$ ways to do one or the other.

Step-by-Step Solution

Step 1: Define variables and restate constraints.

Let $I$ be the number of Indians and $F$ be the number of foreigners in the committee. We are given that there are 6 Indians and 8 foreigners available. The constraints are:

1. $I \ge 2$ (at least 2 Indians)
2. $F = 2I$ (double the number of foreigners as Indians)

Step 2: Determine possible values for $I$ and $F$ based on the constraints and availability.

We need to find integer values of $I$ and $F$ that satisfy the constraints and the availability of members. Since $I \ge 2$, let's check possible values of $I$:

• If $I = 2$, then $F = 2(2) = 4$. Since we have 6 Indians and 8 foreigners available, this is a valid case.
• If $I = 3$, then $F = 2(3) = 6$. Since we have 6 Indians and 8 foreigners available, this is a valid case.
• If $I = 4$, then $F = 2(4) = 8$. Since we have 6 Indians and 8 foreigners available, this is a valid case.
• If $I = 5$, then $F = 2(5) = 10$. We only have 8 foreigners available, so this case is not possible.
• If $I > 5$, it's not possible since there are only 6 Indians.

Therefore, the possible compositions of the committee are: (2 Indians, 4 Foreigners), (3 Indians, 6 Foreigners), and (4 Indians, 8 Foreigners).

Step 3: Calculate the number of ways to form a committee with 2 Indians and 4 Foreigners.

The number of ways to choose 2 Indians from 6 is ${}^6C_2 = \frac{6!}{2!4!} = \frac{6 \times 5}{2 \times 1} = 15$. The number of ways to choose 4 Foreigners from 8 is ${}^8C_4 = \frac{8!}{4!4!} = \frac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2 \times 1} = 70$. By the multiplication principle, the total number of ways for this case is $15 \times 70 = 1050$.

Step 4: Calculate the number of ways to form a committee with 3 Indians and 6 Foreigners.

The number of ways to choose 3 Indians from 6 is ${}^6C_3 = \frac{6!}{3!3!} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20$. The number of ways to choose 6 Foreigners from 8 is ${}^8C_6 = \frac{8!}{6!2!} = \frac{8 \times 7}{2 \times 1} = 28$. By the multiplication principle, the total number of ways for this case is $20 \times 28 = 560$.

Step 5: Calculate the number of ways to form a committee with 4 Indians and 8 Foreigners.

The number of ways to choose 4 Indians from 6 is ${}^6C_4 = \frac{6!}{4!2!} = \frac{6 \times 5}{2 \times 1} = 15$. The number of ways to choose 8 Foreigners from 8 is ${}^8C_8 = \frac{8!}{8!0!} = 1$. By the multiplication principle, the total number of ways for this case is $15 \times 1 = 15$.

Step 6: Calculate the total number of ways to form the committee.

Since the cases are mutually exclusive, we use the addition principle to add the number of ways for each case: Total number of ways = $1050 + 560 + 15 = 1625$.

Common Mistakes & Tips

• Remember to check if the calculated number of people to be selected in each category is possible, given the total number available.
• Carefully read and interpret the phrase "at least". It implies greater than or equal to.
• Make sure to use combinations, not permutations, since the order of selection does not matter.

Summary

We analyzed the constraints to determine the possible compositions of the committee. We then calculated the number of ways to form a committee for each possible composition using combinations and the multiplication principle. Finally, we added the number of ways for each case to find the total number of ways to form the committee.

Final Answer

The final answer is \boxed{1625}, which corresponds to option (D).