Question

Three person A,B and C are standing in a queue.there are five person A and B and eight person between B and C.if there are three person of C and 21 before A,then what could be the minimum no. of person in the queue ?

Answer 1

Three persons A, B, C can be arranged in a queue in six different ways i.e., ABC, CBA, BAC, CAB, BCA, ACB. But since there are only 3 persons ahead of C, so C should be in front of the queue. Thus, there are only two possible arrangements i.e., CBA and CAB. We may consider the two cases as under :

                 3         8        5       21
Case I : ←  C  ↔  B  ↔  A  →
clearly, number of persons in the queue = (3 + 1 + 8 + 1 + 5 + 1 + 21) = 40.

                    3                 5
Case II :  ←  C     A  ↔  B
               


Number of persons between A and C = (8 - 6) = 2.
Clearly, number of persons in the queue = (3 + 1 + 2 + 1 + 21) = 28.
Now, 28 < 40. So, 28 is the minimum number of perosn in the queue.

Answer 2

Answer:

Step-by-step explanation:

Three persons A, B, C can be arranged in a queue in six different ways i.e., ABC, CBA, BAC, CAB, BCA, ACB. But since there are only 3 persons ahead of C, so C should be in front of the queue. Thus, there are only two possible arrangements i.e., CBA and CAB. We may consider the two cases as under :

                 3         8        5       21

Case I : ←  C  ↔  B  ↔  A  →

clearly, number of persons in the queue = (3 + 1 + 8 + 1 + 5 + 1 + 21) = 40.

                    3                 5

Case II :  ←  C     A  ↔  B

               

Number of persons between A and C = (8 - 6) = 2.

Clearly, number of persons in the queue = (3 + 1 + 2 + 1 + 21) = 28.

Now, 28 < 40. So, 28 is the minimum number of perosn in the queue