The total number of distinct ways to arrange four persons (P, Q, R, S) in a row is given by the permutation formula $P(n, n) = n!$.
For 4 persons, the total arrangements are $4! = 4 \times 3 \times 2 \times 1 = 24$. Let this be $N$.
We need to find arrangements where two conditions are met simultaneously: 1. P and R are NOT adjacent. 2. S is seated to the RIGHT of Q. We use the inclusion-exclusion principle. Let $A$ be the set of arrangements where P and R ARE adjacent. Let $B$ be the set of arrangements where S is to the LEFT of Q. We want to find the number of arrangements that are neither in $A$ nor in $B$. This is given by $N - |A \cup B| = N - (|A| + |B| - |A \cap B|)$.
Eight students (P, Q, R, S, T, U, V, and W) are playing musical chairs. The figure indicates their order of position at the start of the game. They play the game by moving forward in a circle in the clockwise direction. After the 1st round, 4th student behind P leaves the game. After 2nd round, 5th student behind Q leaves the game. After 3rd round, 3rd student behind V leaves the game. After 4th round, 4th student behind U leaves the game. Who all are left in the game after the 4th round?
Note: The figure shown is representative.
Six persons P, Q, R, S, T and U are sitting around a circular table facing the center not necessarily in the same order. Consider the following statements:
• P sits next to S and T.
• Q sits diametrically opposite to P.
• The shortest distance between S and R is equal to the shortest distance between T and U.
Based on the above statements, Q is a neighbor of
Seven cars P, Q, R, S, T, U and V are parked in a row not necessarily in that order. The cars T and U should be parked next to each other. The cars S and V also should be parked next to each other, whereas P and Q cannot be parked next to each other. Q and S must be parked next to each other. R is parked to the immediate right of V. T is parked to the left of U.
Based on the above statements, the only INCORRECT option given below is: