Counting-Permutation Problem

[bretania]

Here's another one. What's your take on this? I think the answer is E but the OA says A. Please note that E is 3 times of A.

Pedro's license plate contains 4 characters, and the first character is a letter and the remaining three characters are digits. If the letters O, I and Z are not to be used, and if the digit zero cannot be used more than once, how many different license place numbers can be put together?

(A) 18630 (B) 21060 (C) 23000 (D) 26000 (E) 55890

[clock60]

ok let try
first symbol is any letter from 23 (26-3=23 so we cannot use O,I,Z)
second symbol any digit from 10 as we have no restrictions 10 outcomes
third any digit exept 0, 9 outcomes
and the last also 9 out comes
23*10*9*9=18630
the answer is A
hope my approach is wright and can help

[Crack_Mat]

Originally Posted by clock60

ok let try
first symbol is any letter from 23 (26-3=23 so we cannot use O,I,Z)
second symbol any digit from 10 as we have no restrictions 10 outcomes
third any digit exept 0, 9 outcomes
and the last also 9 out comes
23*10*9*9=18630
the answer is A
hope my approach is wright and can help

First of all answer can not be greater than 23000 so E can not be the right choice!

Here is another approach, but i do not get any answer in the given choices.

ABCD is the number:

A can have 23 values (26 - 3).
B,C,D are number (0-9), based on the given condition that 0 can not come more than once.

So combination in which 0 doesn't come = 23*9*9*9
combination in which 0 comes = (23*1*9*9)*3
so total = (23*9*9*9)+(23*3*9*9)
= 23*9*9*(9+3)
= 23*9*9*12 = 22356

[Crack_Mat]

Another approach:
total number possible = 23*10*10*10 = 23000
now numbers in which all digits are 0 = 23*1*1*1 = 23
number in which two digits are 0 = (23*1*1*9 + 23*1*9*1 + 23*9*1*1)
= 23*27
so total numbers in which 0 is more than once = 23 + 23*27 = 23*28 = 644

So desired answer = 23000 - 644 = 22356

Either two of my approach (which leads to same answer) are wrong or the question is wrongly posed.

[clock60]

[quote=Crack_Mat;804766]Another approach:
total number possible = 23*10*10*10 = 23000
now numbers in which all digits are 0 = 23*1*1*1 = 23
number in which two digits are 0 = (23*1*1*9 + 23*1*9*1 + 23*9*1*1)
= 23*27
so total numbers in which 0 is more than once = 23 + 23*27 = 23*28 = 644

So desired answer = 23000 - 644 = 22356

Either two of my approach (which leads to same answer) are wrong or the question is wrongly posed.[/quote
agree with you
i think that my solution is wrong despite it corresponds the answer
any way we need more solutions or checking the prob

[nandini11]

IMO also the answer is C and I agree with clock60's approach .
Clean ,short and precise.
I would have also attempted this question in the same manner .
23*10*9*9=18630.

[Crack_Mat]

Well here is another problem:
1)4 letter number
2) 1st one can be only a letter and only A can be used.
3) Rest of three are numbers from 0, 1, 2 only
4) 0 can be used only once!

Let me write answer, please correct me in wrong:

answer-1:
1*3*2*2 = 12

answer-2
with no zero = 1*2*2*2 = 8
with only one zero = (1*1*2*2)*3 = 12
so answer = 12 + 8 = 20.

Now which one is correct????
Let us write the set of number which can be formed:
A011 A012 A021 A022
A101 A102 A110 A111 A112 A120 A121 A122
A201 A202 A210 A211 A212 A220 A221 A222
total = 20!!

[sannddyy]

23*10*9*9

This can be repeated three times one for each digit thus answer is three times A , E

[Mits83]

Originally Posted by sannddyy

23*10*9*9

This can be repeated three times one for each digit thus answer is three times A , E

yep, agreed.

it's either 23*10*9*9 or 23*9*10*9 or 23*9*9*10

=> 23(3*9*9*10)
=> 55890

[madboy]

I did it my own way and came up with 22356.

So going with Crack Mat.

A B C D

A can have 23 possible values

Now let's consider plates with and without 0 separately.

With zero: Since only one 0 can be used, that 0 can be placed in any of the 3 positions - B,C and D. For each position of 0, the remaining two places can have 9^2 possible values.
Hence, total no of plates with a zero in them = 23*3*81

Without zero: Each of the places B,C and D can be filled with 9 possible values. Hence, total no of plates with no zero in them = 23*9*9*9

Adding, we get 22356