Correct Option: (B) 76
Let, $S = \left \{ 1, 2, 3, 4, 5 \right \}$,
$S_1, S_2, S_3, S_4, S_5$ be the the five integers,
$|S_i|$ where $i \in S$ mean that integer $S_i$ is fixed .
Now, we want to find the no. of permutations in which at least 1 integer is fixed.
In other words, we want to find $|S_1 \cup S_2 \cup S_3 \cup S_4 \cup S_4 \cup S_5|$.
Now, according to the Principle of Arrangement and De-arrangement, we have
$|S_1 \cup S_2 \cup S_3 \cup S_4 \cup S_4 \cup S_5| = \sum_{i \in S}|S_i| - \sum_{i, j \in S \ \land \ i \neq j}|S_i \cap S_j| + \\ \sum_{i, j, k \in S \ \land \ i \neq j \neq k}|S_i \cap S_j \cap S_k| - \sum_{i, j, k, l \in S \ \land \ i \neq j \neq k \neq l}|S_i \cap S_j \cap S_k \cap S_l| + \\ \sum_{(i, j, k, l, m \in S \ \land \ i \neq j \neq k \neq l \neq m}|S_i \cap S_j \cap S_k \cap S_l \cap S_m|$
Now,
$\sum_{i \in S}|S_i| = \binom{5}{1} \times 4! = 120$
$\sum_{i, j \in S \ \land \ i \neq j}|S_i \cap S_j| = \binom{5}{2} \times 3! = 60$
$\sum_{i, j, k \in S \ \land \ i \neq j \neq k}|S_i \cap S_j \cap S_k| = \binom{5}{3} \times 2! = 20$
$\sum_{i, j, k, l \in S \ \land \ i \neq j \neq k \neq l}|S_i \cap S_j \cap S_k \cap S_l| = \binom{5}{4} \times 1! = 5$
$\sum_{(i, j, k, l, m \in S \ \land \ i \neq j \neq k \neq l \neq m}|S_i \cap S_j \cap S_k \cap S_l \cap S_m| = \binom{5}{5} = 1$
$\therefore$ We finally have,
$|S_1 \cup S_2 \cup S_3 \cup S_4 \cup S_4 \cup S_5| = 120 - 60 + 20 - 5 +1 = 76$
