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n ( > 4 ) kids are standing in a line one next to the other, such that everybody is facing north. How many kids are there? Statement 1: Number of ways in which 3 kids can be selected such that no two selected kids are standing next to each other is 10. Statement 2: If the kids are made to sit around a circular table such that one particular kid always occupies a specific seat, the number of such arrangements possible is 120.
n ( > 4 ) kids are standing in a line one next to the other, such that everybody is facing north. How many kids are there? Statement 1: Number of ways in which 3 kids can be selected such that no two selected kids are standing next to each other is 10. Statement 2: If the kids are made to sit around a circular table such that one particular kid always occupies a specific seat, the number of such arrangements possible is 120.
It is powerplay quesn. In this question's answer, the answer given form 1st statement is : From statement 1: Let there be n kids in the line. The number of ways of selecting 3 of them such that the selected kids are not next to each other is n−2C3=10 => n-2 = 5 => n = 7. Can you elaborate ?
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Hi Amit,
There are 'n' kids in the line, out of which 3 are selected. Let the arrangement be represented as follows:
a 1 b 1 c 1 d
In this, '1' represents a selected kid. 'a' represents the number of kids to the left of the first selected kid, 'b' represents the number of kids between the first two selected kids and so on.
From the condition given in the qn, b and c should be greater than or equal to 1. On the other hand, a and d can take 0 as well.
So, the equation is a + b + c + d = n-3, where b, c >= 1
This is the same as a+b+c+d = n-5, where a,b,c,d >= 0
The answer to this is $$^{n-5+4-1}C_{4-1}$$ = $$^{n-2}C_3$$ = 10 => n = 7
