There are 8 boys of different heights in how many ways can boys can 8 boysbe arranged so that the tallest and shortest
Answers
Answer:
Step-by-step explanation:
Eight people of different heights are to be seated in a row. The shortest and tallest in this group are not seated at either end. What is the probability that:
(a) The tallest and shortest persons are sitting next to each other?
(b) There is one person sitting between the tallest and shortest?
My approach to (a) goes like that: the probability should be equal to
P=6∗5∗5!∗2!8!−7!−7!+2!6!
The reason for the denominator is because I subtract from the total number of combinations those combinations where the tallest is at one end (he/she is "fixed", the rest can arrange in 7! ways, the same goes for the shortest, finally I add 2!6! to avoid double counting. The reason for the numerator is because if the tallest and shortest can't be at the ends, then, since there is eight people in total, one end can be occupied in 6 ways and the other end in 5 ways. If the tallest and shortest are to be sitting together, then instead of 6!, I multiply by 5!2!. My textbook says the answer to (a) is 528,
Given : There are 8boys of different heights.
To Find : in how many ways can 8boys be arranged so that the tallest and shortest boys are not together
Solution:
8 boys can be arranged in 8! Ways
8!
Assume that tallest and shortest boys are together consider them as 1 who can be arranged in 2 ways
now 1 this group and 6 remaining
Hence total = 7
Can be arranged in 7!.2 Ways
Ways 8 boys can be arranged so that the tallest and shortest boys are not together = 8! - 7!.2
= 7! ( 8 - 2)
= 7! .6
= 30240
30240 Ways 8 boys be arranged so that the tallest and shortest boys are not together
Learn More:
In how many ways can the letters of the word "COMPUTER" be ...
https://brainly.in/question/14235713
There are 14 girls and 12 boys all with distinct heights, what are the ...
https://brainly.in/question/12859641