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Rohit Gupta

· started a discussion

· 1 Months ago

Please explain (Question: 195602.0 )

Please explain 15 kahaan se aaya?

Question:
There are 5 persons between A and B. There are 8 persons between B and C. If 3 persons are ahead of C and 21 persons are behind A, then find the maximum and minimum number of persons in the row.
Options:
A) 40 and 28 
B) 28 and 40 
C) 38 and 25 
D) None of these 
Solution:
Ans: (a)


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 person in the queue.

Knowledge Expert

· commented

· 1 Months ago

Dear student,
in the II case , since 8 persons between B and C and 5 persons between A and B we can arrange B,A & C in the following manner
B_ _ _ _ _A_ _ _ C and 21 persons are behind A so we arrange those 21 persons _ _ _15_ _ B_ _ _5 _ _A_ _ _ C
Therefore 15+ 1(B)+5 =21 persons are behind A.
i hope the idea is clear
Thank you
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