P, Q, R, S और T पाँच व्यक्ति हैं, जिनमें से प्रत्येक व्यक्ति को एक कार्य सौंपना है। कार्य-1 न तो P को, न ही Q को सौंपा जा सकता है। कार्य-2 या तो R को, या S को ही सौंपा जाना है। कार्य कितने प्रकार से सौंपे जा सकते हैं ?
सही उत्तर (A) 6 है।
व्याख्या:
There are five persons P, Q, R, S, T and 5 tasks Task-1, Task-2, Task-3, Task-4, Task-5. Each person must be assigned exactly one task. Constraints: 1. Neither P nor Q can be assigned Task-1. This means Task-1 must be assigned to one of the remaining 3 persons (R, S, or T). So there are 3 choices for Task-1. 2. Task-2 must be assigned to either R or S. So there are 2 choices for Task-2. Let us calculate step by step: Case A: Suppose Task-1 is assigned to T (1 choice). Then Task-2 must be assigned to either R or S (2 choices). After assigning Task-1 and Task-2, 3 persons and 3 tasks remain, which can be assigned in 3! = 6 ways. Total for this case = 1 (for T) × 2 (for Task-2) × 6 = 12 ways. Case B: Suppose Task-1 is assigned to R (1 choice). Then R is already used. Task-2 must be assigned to either R or S. But R cannot take Task-2 because R has Task-1! So Task-2 MUST be assigned to S (1 choice). After assigning Task-1 to R and Task-2 to S, 3 persons (P, Q, T) and 3 tasks remain. But remember P and Q cannot take Task-1 (already assigned to R), but for the remaining 3 tasks, let's count valid permutations: Total ways to assign remaining 3 tasks to P, Q, T is 3! = 6 ways. But we must subtract cases where Q or P violates any constraint (none violate since Task-1 is already filled). Wait, let's do it more systematically using inclusion-exclusion or direct assignment: Total persons = 5. Task-1 can be done by R, S, T (3 choices). After assigning Task-1, let's assign Task-2, which can be done by R or S. To avoid double counting the person doing Task-1 and Task-2: Sub-case 1: Task-1 is given to T. Task-2 can be given to R or S (2 choices). Remaining 3 persons and 3 tasks can be arranged in 3! = 6 ways. Total = 1 × 2 × 6 = 12 ways. Sub-case 2: Task-1 is given to R. Task-2 MUST be given to S (1 choice, since R has Task-1). Remaining 3 persons (P, Q, T) and 3 remaining tasks can be arranged in 3! = 6 ways. Total = 1 × 1 × 6 = 6 ways. Sub-case 3: Task-1 is given to S. Task-2 MUST be given to R (1 choice, since S has Task-1). Remaining 3 persons and 3 remaining tasks can be arranged in 3! = 6 ways. Total = 1 × 1 × 6 = 6 ways. Total valid assignments = 12 + 6 + 6 = 24 ways? Wait, let's re-verify: Total permutations of 5 persons to 5 tasks = 5! = 120. Violations: P gets Task-1 (4! = 24 ways), Q gets Task-1 (4! = 24 ways). Both P and Q get Task-1 is impossible. So invalid assignments where Task-1 goes to P or Q = 24 + 24 = 48 ways. Valid assignments for Task-1 = 120 - 48 = 72 ways. Now among these 72 ways, we must satisfy the condition that Task-2 goes to R or S. Let's count how many of the 72 assignments have Task-2 given to R or S. Total ways where Task-1 not P,Q is 72. Task-2 can be assigned to any of the remaining 4 persons. By symmetry, exactly 2 out of 4 available persons are R or S. Thus, 2/4 of the valid 72 assignments have Task-2 given to R or S. So 72 × (2/4) = 72 × (1/2) = 36 ways? Wait, let's re-check options: 6, 12, 18, 24. Let's re-evaluate Sub-case 1, 2, 3: 12 + 6 + 6 = 24 ways. Therefore, option (d) 24 is correct.
In English (Question & Model Answer)
There are five persons P, Q, R, S and T each one of whom has to be assigned one task. Neither P nor Q can be assigned Task-1. Task-2 must be assigned to either R or S. In how many ways can the assignment be done ?
The correct answer is (A) 6.
Explanation:
There are five persons P, Q, R, S, T and 5 tasks Task-1, Task-2, Task-3, Task-4, Task-5. Each person must be assigned exactly one task. Constraints: 1. Neither P nor Q can be assigned Task-1. This means Task-1 must be assigned to one of the remaining 3 persons (R, S, or T). So there are 3 choices for Task-1. 2. Task-2 must be assigned to either R or S. So there are 2 choices for Task-2. Let us calculate step by step: Case A: Suppose Task-1 is assigned to T (1 choice). Then Task-2 must be assigned to either R or S (2 choices). After assigning Task-1 and Task-2, 3 persons and 3 tasks remain, which can be assigned in 3! = 6 ways. Total for this case = 1 (for T) × 2 (for Task-2) × 6 = 12 ways. Case B: Suppose Task-1 is assigned to R (1 choice). Then R is already used. Task-2 must be assigned to either R or S. But R cannot take Task-2 because R has Task-1! So Task-2 MUST be assigned to S (1 choice). After assigning Task-1 to R and Task-2 to S, 3 persons (P, Q, T) and 3 tasks remain. But remember P and Q cannot take Task-1 (already assigned to R), but for the remaining 3 tasks, let's count valid permutations: Total ways to assign remaining 3 tasks to P, Q, T is 3! = 6 ways. But we must subtract cases where Q or P violates any constraint (none violate since Task-1 is already filled). Wait, let's do it more systematically using inclusion-exclusion or direct assignment: Total persons = 5. Task-1 can be done by R, S, T (3 choices). After assigning Task-1, let's assign Task-2, which can be done by R or S. To avoid double counting the person doing Task-1 and Task-2: Sub-case 1: Task-1 is given to T. Task-2 can be given to R or S (2 choices). Remaining 3 persons and 3 tasks can be arranged in 3! = 6 ways. Total = 1 × 2 × 6 = 12 ways. Sub-case 2: Task-1 is given to R. Task-2 MUST be given to S (1 choice, since R has Task-1). Remaining 3 persons (P, Q, T) and 3 remaining tasks can be arranged in 3! = 6 ways. Total = 1 × 1 × 6 = 6 ways. Sub-case 3: Task-1 is given to S. Task-2 MUST be given to R (1 choice, since S has Task-1). Remaining 3 persons and 3 remaining tasks can be arranged in 3! = 6 ways. Total = 1 × 1 × 6 = 6 ways. Total valid assignments = 12 + 6 + 6 = 24 ways? Wait, let's re-verify: Total permutations of 5 persons to 5 tasks = 5! = 120. Violations: P gets Task-1 (4! = 24 ways), Q gets Task-1 (4! = 24 ways). Both P and Q get Task-1 is impossible. So invalid assignments where Task-1 goes to P or Q = 24 + 24 = 48 ways. Valid assignments for Task-1 = 120 - 48 = 72 ways. Now among these 72 ways, we must satisfy the condition that Task-2 goes to R or S. Let's count how many of the 72 assignments have Task-2 given to R or S. Total ways where Task-1 not P,Q is 72. Task-2 can be assigned to any of the remaining 4 persons. By symmetry, exactly 2 out of 4 available persons are R or S. Thus, 2/4 of the valid 72 assignments have Task-2 given to R or S. So 72 × (2/4) = 72 × (1/2) = 36 ways? Wait, let's re-check options: 6, 12, 18, 24. Let's re-evaluate Sub-case 1, 2, 3: 12 + 6 + 6 = 24 ways. Therefore, option (d) 24 is correct.