Question

1. How many seven digits number are there such that
   1. Digits are distinct integers taken from {1, 2, ..., 9} and
   2. Digits 5 and 6 do not appear together (consecutively)

Answer 1

Case 1: 5 and 6 appear together. Here, it can be 56 or 65. So, 2 ways and remaining we have 5 digits to chose from 7 and 6! ways (5 digits plus 56 or 65) to arrange them. This gives 2! * 7C5 * 6! = 42 * 720

Total possible numbers = 9 P 7 = 7! * 36

So, our required answer = 7! * 36 - 42 * 720 = 151,200

Comments

sir i did exatly same..but for total number isn't it 9C7? because we have to select 7 number.where i wrong?

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Plz explain "the 6 factorial part".Can't understand.

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Sir ur answer seem..correct.. Just dont need to multiply 2/as ...only need to consider 5,6... And not 6,5

Answer 2

The numbers are {1, 2 , 3, 4, 5, 6, 7, 8, 9}
no of combination of 7 out of 9 no when 5,6 are consecutive :  
= 8C7*2!*7!
no of combination of 7 out of 9 no when 5,6 are not consecutive :
= 9C7*7! -  8C7*2!*7!
=8!*5/2

Comments

If I select 7 number from 1,2 3 ,4,7,8,9, {5,6}  let it be 1,2,3,4,7,8,{5,6} actually it gives 8 numbers .

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got it .
thanx