The binomial theorem is used to expand expressions of the form: a) (a + b) b) (a + b) c) (a + b)n d) (a - b)2 e) None of the above.

We need to select 4 people from 8 candidates for 4 distinct positions.

Here, $n=8$ (total candidates) and $k=4$ (positions to fill).

• The correct option is e) None of the above. (The calculated answer is 1680, which is not among options a, b, c, d).

50. How many different 7-letter words can be formed using the letters of the word "INSTITUTE"?

The word "INSTITUTE" has 9 letters: I (2 times), N (1 time), S (1 time), T (3 times), U (1 time), E (1 time).
We need to form 7-letter words using these letters. This requires considering different cases based on the repetitions of 'I' and 'T'.

Case 1: Using three T's, two I's, and two other distinct letters.

• Choose 2 distinct letters from {N, S, U, E}: $\binom{4}{2} = 6$ ways.
• Arrange the 7 letters (T, T, T, I, I, X, Y): $\frac{7!}{3!2!} = \frac{5040}{6 \times 2} = 420$ ways.
• Total for Case 1: $6 \times 420 = 2520$.

Case 2: Using three T's, one I, and three other distinct letters.

• Choose 3 distinct letters from {N, S, U, E}: $\binom{4}{3} = 4$ ways.
• Arrange the 7 letters (T, T, T, I, X, Y, Z): $\frac{7!}{3!1!} = \frac{5040}{6} = 840$ ways.
• Total for Case 2: $4 \times 840 = 3360$.

Case 3: Using two T's, two I's, and three other distinct letters.

• Choose 3 distinct letters from {N, S, U, E}: $\binom{4}{3} = 4$ ways.
• Arrange the 7 letters (T, T, I, I, X, Y, Z): $\frac{7!}{2!2!} = \frac{5040}{4} = 1260$ ways.
• Total for Case 3: $4 \times 1260 = 5040$.

Case 4: Using two T's, one I, and four other distinct letters.

• Choose 4 distinct letters from {N, S, U, E}: $\binom{4}{4} = 1$ way.
• Arrange the 7 letters (T, T, I, N, S, U, E): $\frac{7!}{2!1!} = \frac{5040}{2} = 2520$ ways.
• Total for Case 4: $1 \times 2520 = 2520$.

Case 5: Using one T, two I's, and four other distinct letters.

• Choose 4 distinct letters from {N, S, U, E}: $\binom{4}{4} = 1$ way.
• Arrangement: $\frac{7!}{2!1!} = \frac{5040}{2} = 2520$ ways.
• Total for Case 5: $1 \times 2520 = 2520$.

Total number of ways = $2520 + 3360 + 5040 + 2520 + 2520 = 15960$.

• The correct option is e) None of the above. (The calculated answer is 15960, which is not among options a, b, c, d).

51. The binomial theorem is used to expand expressions of the form:
The binomial theorem provides a formula for expanding algebraic expressions of the form $(a+b)^n$ for any non-negative integer $n$.

• The correct option is c) $(a+b)^n$.

52. The coefficient of the middle term in the expansion of $(x+y)^n$ is given by:

If $n$ is an even number, there is one middle term, which is the $(\frac{n}{2} + 1)$-th term. Its coefficient is $\binom{n}{n/2}$.
If $n$ is an odd number, there are two middle terms, the $(\frac{n+1}{2})$-th and $(\frac{n+3}{2})$-th terms. Their coefficients are $\binom{n}{(n-1)/2}$ and $\binom{n}{(n+1)/2}$, which are equal.

None of the options (a) $\binom{n}{2}$, (b) $n$, (c) $\binom{n+1}{2}$, or (d) $\binom{n-1}{2}$ represent the general formula for the coefficient of the middle term(s).

• The correct option is e) None of the above.

53. The expansion of $(a+b)^4$ is:

For $n=4$:

$\binom{4}{0} = 1$
$\binom{4}{1} = 4$
$\binom{4}{2} = \frac{4 \times 3}{2 \times 1} = 6$
$\binom{4}{3} = 4$
$\binom{4}{4} = 1$

• The correct option is a) $a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + b^4$.

54. The binomial coefficient $\binom{n}{k}$ is given by:
The formula for the binomial coefficient $\binom{n}{k}$ is defined as the number of ways to choose $k$ elements from a set of $n$ elements without regard to the order of selection.

• The correct option is d) $n! / (k!(n-k)!)$.

55. The binomial coefficient (7 choose 3) is:

(7 choose 3) is $\binom{7}{3}$.

Here, $n=7$ and $k=3$.

• The correct option is b) 35.