How many number plates are possible, which contain 3 alphabets followed by 3 numbers from 0 to 9?

M4maths Previous Puzzles

30 Sep 2010 Possible number

How many number plates are possible, which contain 3 alphabets followed by 3 numbers from 0 to 9?
000 is not allowed in any number plate.

OPtion

1) 676
2) 1752
3) 19234524
4) 6
5) 34256
6) 3450
7) 456
8) 17558424
9) 17576
10) 17576000

Solution

Total number plates = 26*26*26*10*10*10 = 17576000
Number plates containing 000 = 26*26*26 = 17576
Hence required number plates = 17576000 – 17576

Correct Option: 8 Best Solution (13)

Dr.Dipin Singh   14 years AGO

(26*26*26* 999)= 17558424 number plates are possible, which contain 3 alphabets followed by 3 numbers from 0 to 9.

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Karthik Pandiyan   14 years AGO

3 alphabets 3 numbers
26*26*26*10*10*10=17576000

000 is not allowed
3 alphabets 000
26*26*26*1*1*1=17576

Hence ans: 17576000-17576=17558424

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Bankim   14 years AGO

(26^3)*999

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S. Singh   14 years AGO

(1000-1)*26*26*26 = 999*17576= 17558424

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Narayanan   14 years AGO

soln: (26*26*26*10*10*10)-(26*26*26) -> 17558424.option: 8

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SUDESHNA SEN   14 years AGO

First 3 places can be filled up by combination of the 26 alphabets in 26^3 ways and next 3 places can be filled up by combination of the 10 numbers (0-9)in 10^3 ways.
So, the total number of combination is 26^3*10^3,of which 26^3 has 000 combination.
So, the required number is= 26^3*10^3-26^3= 26^3(10^3-1)= 17576*999= 17558424..... Option 8)

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Achelal Sharma   14 years AGO

the total number of combination is 26^3 * 10^3 => 17576000
out of which 000 is not allowed which occurs 17576 times
so the total number of combination is 17576000 - 17576 => 17558424

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ASHOK ROY   14 years AGO

First 3 places can be filled up by combination of the 26 alphabets in 26^3 ways and next 3 places can be filled up by combination of the 10 numbers (0-9)in 10^3 ways.
So, the total number of combination is 26^3*10^3,of which 26^3 has 000 combination.
So, the required number is= 26^3*10^3-26^3= 26^3(10^3-1)= 17576*999= 17558424..... Option 8)

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sanjay   14 years AGO

First 3 places can be filled up by combination of the 26 alphabets in 26^3 ways and next 3 places can be filled up by combination of the 10 numbers (0-9)in 10^3 ways.
So, the total number of combination is 26^3*10^3,of which 26^3 has 000 combination.
So, the required number is= 26^3*10^3-26^3= 26^3(10^3-1)= 17576*999= 17558424..... Option 8)

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Mahesh   14 years AGO

17576000-17576 = 17558424

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SARBANI DAS GHOSH   14 years AGO

First 3 places can be filled up by combination of the 26 alphabets in 26^3 ways and next 3 places can be filled up by combination of the 10 numbers (0-9)in 10^3 ways.
So, the total number of combination is 26^3*10^3,of which 26^3 has 000 combination.
So, the required number is= 26^3*10^3-26^3= 26^3(10^3-1)= 17576*999= 17558424..... Option 8)

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sanjib biswas   14 years AGO

First 3 places can be filled up by combination of the 26 alphabets in 26^3 ways and next 3 places can be filled up by combination of the 10 numbers (0-9)in 10^3 ways.
So, the total number of combination is 26^3*10^3,of which 26^3 has 000 combination.
So, the required number is= 26^3*10^3-26^3= 26^3(10^3-1)= 17576*999= 17558424..... Option 8)

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goutam Choudhury   14 years AGO

4 letters can be choosen from 26 letters by 26^3 way. 3 digits on the same way can be choosen from 10 digits by 10^3 ways. as 000 is not allowed therefore the possible numbers is 26^3*(10^3 -1)=17558424. option 8

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