Password

[cartera]

A four character password consists of one letter of the alphabet and three different digits between 0 and 9, inclusive. The letter must appear as the 2nd or 3rd character of the password. How many different passwords are possible?

 

 

A. 5040

B. 18720

C 26000

D 37440

E. 52000

 

 

OA

D

 

I got

E: 2*(10C1^3*26C1)

where am I wrong?

[12rk34]

First & fourth characters can selected in 10 ways each.

If second is a number it can be selected in 10 ways.

If third is alphabet it can be selected in 26 ways.

So total selections (i.e. no. of passwords) = 10^3 x 26 = 26000

Similarly if second character is a alphabet no. of selections will be 26000.

Hence total no. of passwords = 26000 x 2 =52000

Hence E. :)

[rajtitan]

If 2nd apace is a character then the combination would be 10 * 26* 8* 9 =18720

 

similarly if the third space is a character we get the same 18720 ways

 

So in total its 37440 , its Option D

[12rk34]

Sorry I missed it. Since the digits are all different, if second letter is alphabet then total combinations will be = 10*26*9*8 = 18720

Similarly 18720 will be combinations if third character is alphabet.

Hence total 18720*2 = 37440

[nshabde]

There are two arrangements as stated

 

_ L _ _

or

_ _ L _

 

Digits are not to be repeated, so

 

(10*26*9*8) + (10*9*26*8) = 37440.

 

IMO D

[wawan_andrean]

There are two arrangements as stated

 

_ L _ _

or

_ _ L _

 

Digits are not to be repeated, so

 

(10*26*9*8) + (10*9*26*8) = 37440.

 

IMO D

 

splendid and i absolutely agree with your approach, buddy.

since digit number is unique therefore 3 out of 4 slots should be 10*9*8 and the last one slot is belongs to the one unique letter, total 26. multiply all of them resulted 18720. Since there are two scenario for letter is placed, 18720+18720=37440

[genius_in_the_gene]

2*(10*26*9*8)