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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?

  1. $164025$
  2. $190951$
  3. $194976$
  4. $219049$
in Combinatory by Loyal (7.2k points)
edited by | 360 views

2 Answers

+4 votes
Best answer
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.
by Loyal (7.8k points)
selected by
+1
Final statement is correct?
0
Edited :)
+1 vote
$n(A)$ = 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
by Active (2k points)
0
(5^2-4^2) * ( 10^4 - 9^4) = 30951 :)
+2
yeah now i wonder how i got 59049 by calculation. Concept is same i guess :)

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