Company X ordered for security codes to be formed for each of its empl

You can pick digits until 3 ways and alphabets in 2 ways

Therefore 2 digits can be picked in 3*3 ways and 3 alphabets in 2*2*2 = 72 ways

These 5 characters can be arranged in 5c2 or 5C3 ways since once you pick where the digits or alphabets have to go the remainder will go in the remaining places.

So 72x10= 720

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neelugarg33
There are two things which are required to be done here arrangement of letters which can be done in 5!/2!3!
And choosing digits and characters with repetition 3*3*2*2*2
Hope this helps



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Solution:

Since the letters and digits can be repeated, the 3 letters can be arranged in 2 x 2 x 2 = 8 ways and 2 digits can be arranged in 3 x 3 = 9 ways and if L represents a letter and D represents a digit, then LLLDD can be arranged in 5!/(3!2!) = 10 ways. Therefore, there are 8 x 9 x 10 = 720 different codes possible.

Answer: D

Each code has 5 characters

Each code will have 3 of these characters as letters (X or Y) and 2 of these characters as Digits (1 or 2 or 3)

(1st) out of the 5 character order, choose which 3 spaces the Letters will take and which 2 spaces the Digits will take

For instance we can have:

Letter - letter - letter - digit - digit

Or

Lettter - letter - digit - letter - digit

Etc.

The number of different unique arrangements is given by: 5! / 3! 2! = 10

AND

(2nd) for each one of these 10 arrangements, we need to choose from the available options of Digits and Characters. Repeatition is allowed.

Digit 1 - 3 available options

Digit 2 - 3 available options

Letter 1 - 2 available options

Letter 2 - 2 available options

Letter 3 - 2 available options


Applying the Factor Foundation Rule, the number of unique codes that can be formed is:

(10) * (3 * 3 * 2 * 2 * 2) =

(10) (9) (8) = 720 possible codes



720

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