Thread: Permutation (Basic) Q2

Hey nikconsult.

Hint: You can arrange the two shortest in two ways and the rest can be arranged using normal permutations.

In other words, you have 5! for one arrangement of the two shortest (A then B) and also for another arrangement (B then A) if A and B are the two smallest.

Because there are no restrictions on the other five people, then it means that they can be represented in 5! or 120 ways for one particular arrangement of the two smallest people.

Note also that the arrangement of the five tallest and the two smallest is independent so you can vary both individually. In the case of independence the number of combinations is C*D where C is the number of possibilities for the first arrangement and D is the number for the second arrangement.

Dear Chiro & All

it mean the two smallest will be at box no 6 & 7 and it can be interchange?

Actually, I take exception to the statement of the problem "the two shortest are at the end lf the line". The line has two ends. They must mean "at the ends of the line"- one on the left of the line, the other on the right.

First, take the two shortest students and move them aside. Order the other 5 anyway you want. There are 5!= 120 ways to do that. Choose one of the two shortest students to be on the left. There are two ways to do that. Put the last student on the right. There are a total of (120)(2)= 240 ways to order the students.

(If they mean that one end of the line is, say, against a wall and they want to put the two shortest on the other end, we have the same result. there are 5!= 120 ways to order the first 5 students, then 2 ways to choose one of the two shortest to be next, then only one student to put at the end of the line. 2(120)= 240 ways to order them.)

i think i already understand..i will post my calculation for your comments.