Question

How many number plates can be made if the number plates have two letters of the english alphabet (a-z) followed by two digits (0-9) if the repetition of digits or alphabets is not allowed?

a. 56800

b. 56500

c. 52500

d. 58500?

Answer 1

Option d. is the answer

26 * 25 * 10 * 9 = 58500
Explanation:
[ 2- letters of Alphabets from (a-z)] and 2 digits from (1-9). ****No repetitions are allowed in any case]

Finding no. of ways of selecting 2 letters from a-z without repetition

Select any alphabet first.
26C1 = 26
now 25 Alphabets left,
(As another alphabet should not be equal to the one that is selected)
taking one from these 25
25C1=25

Finding no. of ways of selecting 2 digits from (0-9) without repetition:

Total 10 no.s are there in 0-9
First taking one of them,
10C1=10
now 9 no.s left
(The other should not be equal to this one)
9C1= 9

As all are ''and''ed multiply all,
I mean selecting one from the alphabets AND other one from alphabets left AND selecting one digit from 0-9 AND other digit from remaining digits....that is..
26*25*10*9=58500

Hope it helps
;)
Mark brainliest if you find this helpful..

Answer 2

Given: The number plates have two letters of the english alphabet (a-z) followed by two digits $(0-9)$.

To Find: The number of plates that can be made if the repetition of digits or alphabets are not allowed.

Solution:

Understand that there are $26$ letters in an alphabet and select two alphabets without repetition.

So, the total number of ways$\[{ = ^{26}}{C_1}{ \times ^{25}}{C_1}\]$

Similarly, there are $(0-9)$ digits and select two digits without repetition.

So, the total number of ways$\[{ = ^{10}}{C_1}{ \times ^9}{C_1}\]$

Therefore, the total number of plates that can be made if the repetition of digits or alphabets is not allowed is

$\[\begin{array}{l}^{26}{C_1}{ \times ^{25}}{C_1}{ \times ^{10}}{C_1}{ \times ^9}{C_1}\\ = 26 \times 25 \times 10 \times 9\\ = 650 \times 90\\ = 58500\end{array}\]$

Hence, the correct option is (d) i.e. $58500$.

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