Probability MCQ3: (No.of ways for Arrangement)
Probability MCQ3: What are the chances that no two boys are sitting together for a photograph if there are 5 girls and 2 boys?
a) 1/21 b) 4/7 c) 2/7 d) 5/7
Solution:
Keep in mind that :
If there are n things, then number of ways to arrange them = n!
Where,
\[ n!=n \times (n-1) \times (n-2) \times (n-3) —– so – on \]
( we read n! as “n factorial” —a mathematical terminology for a number n !)
Let us first understand factorial concept by examples;
What is 4! ?
Ans: \( 4!=4\times 3\times 2\times1=24 \)
What is 6 !?
ans :\( 6!=6\times 5 \times 4 \times 3 \times2 \times1=720 \)
Now lets come to the question:
Number of girls =5
Number of boys =2
Total=5+2=7
so, number of ways to arrange 7 persons =7! = 7x6x5x4x3x2x1=5040
i-e we can arrange 7 persons in different 5040 ways !
If lets suppose two boys sit together – we take their pair as one person ! – then number of total persons will be= 5+1=6
Then number of ways to arrange 6 persons = 6! = 6x5x4x3x2x1=720
or
number of ways to arrange 5 girls and one pair of boys=720
Since together sitting boys (pair of boys) can also interchange their positions – i-e one boy sitting right side of the other can sit left as well ! — two ways of sitting together- left or right side!
So, the number of ways doubles in this case !
i-e number of ways to arrange 5 girls and two together sitting boys= 2x 720=1440
Now ,
Number of ways of arrangement that no two boys sit together =Total Number of ways of arrangement of 7 person – The number of ways in which two boys sit together = 5040 – 1440 = 3600
We know for Probability of an ‘event’, formula in statistics is;
\[ P(Event)=\frac {Event}{TotalSample} \]
Here the event is “number of ways boys not sit together” – and total sample is “number of ways of arranging 7 person”
So ,
if we suppose,
E=number of ways boys not sit together
and
T=number of ways of arranging 7 person
then,
\[ P(E)=\frac {3600}{5040}=\frac {5}{7} \]
So the option ‘d’ is correct !
