in how many ways can 4 Indians and 3 Americans be seated in a row such that no two Americans are seated together?
Answers
Given : 4 Indians and 3 Americans in a row
To Find : Number of ways no two Americans are seated together
Solution:
4 Indians
3 American (A₁ , A₂ , A₃)
Total = 4 + 3 = 7
7 can be arranged in 7 ! = 5040 Ways
cases when 2 Americans are seating together
2 Americans out of 3 can be selected in ³C₂ = 3 ways
and can be arranged in 2! = 2 Way
A group of 2 Americans + 5 ( 1 American + 4 Indians) = 6
6 can be arranged in 6 ! ways
Total Ways 2 Americans are seating together 3 x 2 x 6!
= 4320 Ways
But this case Involve Duplicate cases like
(A₁ , A₂) ,A₃ = A₁ , (A₂ , A₃)
same configuration counted 2 times
So we need to subtract case of 3 Americans together from this to get Actual cases when at least 2 Americans are seating together
3 Americans out of 3 can be selected in ³C₃ = 1 way
and can be arranged in 3! = 6 Way
A group of 3 Americans + 4 Indians = 5
5 can be arranged in 5 ! ways
Total Ways 3 Americans are seating together 6 x 5!
=720 Ways
no two Americans are seated together = 5040 - 4320 + 720
= 1440
in 1440 ways 4 Indians and 3 Americans be seated in a row such that no two Americans are seated together
Another method : Simpler one
4 Indian can seat in 4 ! = 24 Ways (no restrictions )
now 3 Americans has 5 spaces in total to sit where they has to choose 3
- I - I - I - I -
spaces that can be done in ⁵C₃ = 10 ways
and 3 American can be arranged in 3! = 6 ways
Total Ways = 24 x 10 x 6 = 1440 Ways
in 1440 ways 4 Indians and 3 Americans be seated in a row such that no two Americans are seated together
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