Question

How many 4 digit integers are there with digit 6 appearing exactly once .

6_ _ _   9 *8*7
_ 6 _ _  9* 8 * 7

_ _ 6 _  9*9*7

_ _ _ 6  9*9*8

I'm getting as 2223 .
Is my approach correct . In book I'm getting a different ans .

Answer 1

6_ _ _   = 9*9*9

_ 6 _ _  = 8*9*9

_ _ 6 _   = 8*9*9

_  _ _6   = 8*9*9

total numbers = (9+8*3)*9*9 = 33*9*9 = 2673

Answer 2

6_ _ _  :9*8*7=504

_6_ _ :  8*8*7=448 ( we can't take 0 at first place therefore the first place will have only 8 choices and for the next place we will have 8 choices because 0 has been included and for the next 7 choices )

_ _6 _: 8*8*7=448

_ _ _6: 8*8*7=448

Total=504+448+448+448=1848

Comments

- 6 - - = 8*8*7 ?
how....?

0-9 = 10 digits are there isn't it

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yes its 9 digit but number starts with 0 will be of 3 digit for example 0615 it is of 3 digit only

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it has to be 9*8*7
[0-9 ] =9 digits possible in in 1000th position

6 = fixed in 100th pos

[0-9] (minus) digit used in 1000th position and then 6 = 8 digits osible in 10th position

Simlarly 7 digits possible in unit position

 

SO ,  9*8*7 ...
IS it NOT !?

You said 8*8*7 . How this is possible ? I didn't get your answer

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you mean.... other digits won't repeat?