ISI2019-MMA-20

Question

Suppose that the number plate of a vehicle contains two vowels followed by four digits. However, to avoid confusion, the letter $‘O’$ and the digit $‘0’$ are not used in the same number plate. How many such number plates can be formed?

• A. $164025$
• B. $190951$
• C. $194976$
• D. $219049$

Answer 1

It is easy to solve this by the principle of complement.

The total number of number-plates CONTAINING both O and 0 are:

5*5 ways to fill the first two positions, of which 4*4 picks won't have the letter O. So, 5*5 - 4*4 = 9 arrangements will have the letter O.

10*10*10*10 ways to fill rest four positions, of which 9*9*9*9 arrangements won't have the digit 0. So, 10000-6561= 3439 arrangements will have the digit 0.

So, 9 * 3439 arrangements will have both O and 0.
Hence 5*5*10*10*10*10  -  9 * 3439 = 219,049 arrangements will NOT CONTAIN both O and 0.

Comments

Final statement is correct?

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Edited :)

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can anyone tell me why we can't direct sove like we have 4 choice for vowels and 9 choice for the number so

4*4*9*9*9*9=104976 . can't we do lie that and if not then what is the mistake which i did in this explanation , please explain. thank you

Answer 2

$n(A)$ = all possible number plates = $5*5*10*10*10*10 = 250000$

$n(B)$ = all number plates in which both letter $O$ and digit $0$ are used atleast once = $(5^2-4^2) *(10^4-9^4) = 59049$

$ans = n(A) -n(B) = 190951 $

So $B$ is correct answer

Comments

(5^2-4^2) * ( 10^4 - 9^4) = 30951 :)

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yeah now i wonder how i got 59049 by calculation. Concept is same i guess :)

Answer 3

Answer: D

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We can divide it into two cases.

Case 1: The number plate contains ‘O’ (no zero).

This case can be further subdivided into two cases: 

Case 1a: Contains ‘O’ at both the places: $9^4$.

Case 1b: Contains ‘O’ at exactly one place: $2\times 4\times 9^4$.

Case 2: Does not contain ‘O’ (may contain zero). $4\times 4\times 10^4$

 

Hence, the total cases = $9^4+8\times 9^4+16\times 10^4=9^5+16\times 10^4=219049$