
If 12 persons are seated in a row, the number of ways of selecting 3 persons from them, so that no two of them are seated next to each other.
a) 85
b) 100
c) 120
d) 240
Answer
162k+ views
Hint: First, select three people from the group of twelve, and then arrange the remaining nine seated people so that there is an empty seat before and after each seated person. Then occupy the empty seats with those three people in different combinations.
Formula Used:The number of ways of selecting r things out of n different things is called combinations.
\[n{C_r} = \dfrac{{n!}}{{(n - r)!r!}}\], \[0 \le r \le n\]
Complete step by step solution:First, select 3 persons out of the group of 12 persons.
Number of person to be seated in a row = 12 – 3
= 9
Now, arrange the 9 person seated in a row in such a way such that there is an empty seat before and after each seated person, as shown below. Yellow shaded boxes represent the seats occupied by 9 persons.
From the figure, it is clear that there are 10 empty seats among and around the 9 seated persons.
The 3 persons who were chosen earlier can now be seated in different combinations among the 10 empty seats, so that no two of them are seated next to each other.
Number of ways 3 persons can be seated among the 10 empty seats = \[10{C_3}\]
\[ = \dfrac{{10!}}{{(10 - 3)!3!}}\]
\[ = \dfrac{{10 \times 9 \times 8}}{{3 \times 2 \times 1}}\]
\[ = 5 \times 3 \times 8\]
= 120
Option ‘C’ is correct
Note: Student may ignore the empty seats at the extremes and calculated wrongly as\[8{C_3}\].
Student may wrongly calculate the empty seats as 9 and calculated wrongly as \[9{C_3}\].
Formula Used:The number of ways of selecting r things out of n different things is called combinations.
\[n{C_r} = \dfrac{{n!}}{{(n - r)!r!}}\], \[0 \le r \le n\]
Complete step by step solution:First, select 3 persons out of the group of 12 persons.
Number of person to be seated in a row = 12 – 3
= 9
Now, arrange the 9 person seated in a row in such a way such that there is an empty seat before and after each seated person, as shown below. Yellow shaded boxes represent the seats occupied by 9 persons.
From the figure, it is clear that there are 10 empty seats among and around the 9 seated persons.
The 3 persons who were chosen earlier can now be seated in different combinations among the 10 empty seats, so that no two of them are seated next to each other.
Number of ways 3 persons can be seated among the 10 empty seats = \[10{C_3}\]
\[ = \dfrac{{10!}}{{(10 - 3)!3!}}\]
\[ = \dfrac{{10 \times 9 \times 8}}{{3 \times 2 \times 1}}\]
\[ = 5 \times 3 \times 8\]
= 120
Option ‘C’ is correct
Note: Student may ignore the empty seats at the extremes and calculated wrongly as\[8{C_3}\].
Student may wrongly calculate the empty seats as 9 and calculated wrongly as \[9{C_3}\].
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