Jayant Kishore
Last Activity: 8 Years ago
Alternatively , you can proceed this way.
No. of elements in sample space = 10 x 9 x 8 for selection of 3 numbers one by one without replacement.
According to the question 3 & 7 must get chosen and the third no. should be between them (ie 4 or5 or 6) as per condition .
So , favourable selection may be as 3 , 7 , X
where X denotes either of 4 or 5 or 6. This can be done in 3 ways.
But , here since we are selecting one by one , IN THIS CASE ORDER OF SELECTION MATTERS
So, no of favourable cases = 3! x 3
Therefore required probablity = 3! x 3/ (10 x 9 x 8) = 1/40.
NOTICE THE DIFFERENCE .
