M4maths Previous Puzzles

02 Jul 2014 Solve

A car plate is written by taking 3 alphabets and 3 integers, if after 3 alphabets 3 zeros are not allowed, then total possible car plates with no code repeated are ?

OPtion

1) 11232000
2) 17576000
3) 17576
4) 17593576
5) 17558424
6) 14236560
7) 15600000
8) 17568424
9) 12812904
10)None of these

Solution

total alphabets are 26 hence total possibilities for taking 3 alphabets are 26*26*26 = 17576
Total integers are 10 hence total possibilities for taking 3 integers are
10*10*10 = 1000
Total number of avoided cases are 26*26*26*1*1*1 = 17576
Therefore answer is 17576 * 1000 - 17576 = 17558424

Option 5)

Correct Option: 5 Best Solution (8)

Ravi Kumar Kodari   10 years AGO

10^3*26^3-26^3
=17558424

Like? Yes (5) | No (1)  

RAKESH   10 years AGO

26*26*26*10*10*10 - 26*26*26*1
= 26^3 * 999
= 17558424

Like? Yes (5) | No (2)  

Ann Theressa   10 years AGO

out of 26 alphabets 3 are taken (26^3)
out of 10 numbers (0 to 9) 3 are taken (10^3)
after 3 alphabets 3 zeros are not allowed.
===> 26^3 out of total combinations is not allowed.

possible combinations = 26^3 * 10^3 - 26^3
= 17576000-17576
=17558424

option : 5

Like? Yes (4) | No (2)  

Ramesh Sriram   10 years AGO

10^3*26^3-26^3*1
=17558424

Like? Yes (3) | No (1)  

Udhaya kumar   10 years AGO

26^3*10^3 - 26^3 = 17558424

Like? Yes (3) | No (1)  

kapil singh   10 years AGO

3 alphabet = 26*26*26=17576
3 integer no. without(000)=999
so, no. of cars plate =17576*999=17558424

Like? Yes (2) | No (1)  

MRITUNJAY KUMAR   10 years AGO

26^3*10^3-(26^3)=17558424

Like? Yes (2) | No (1)  

kavya   10 years AGO

26 alphabets in 3 aggangements so (26)^3
10 integers in 3 (10)^3-1 since 000 case eliminated so both multiplied together gives 17558424

Like? Yes (1) | No