Question

A password is 4 characters long and must consist of 3 letters and 1 of 10 special characters. If letters can be repeated and the special character is at the end of the password, how many possibilities are there? a. 175,760 b. 456,976 c. 703,040 d. 1,679,616 Please select the best answer from the choices provided A B C D.

Answer 1

The number of possibilities for constructing the 4 characters long password with specified conditions is given by: Option C: 703,040

What is the rule of product in combinatorics?

If a work A can be done in p ways, and another work B can be done in q ways, then both A and B can be done in $p \times q$ ways.

Remember that this count doesn't differentiate between order of doing A first or B first then doing other work after the first work.

Thus, doing A then B is considered same as doing B then A

We're specified that:

• The password needs to be 4 characters long
• It must have 3 letters and 1 of 10 special characters.
• Repetition is allowed.

So, each of 3 characters get 26 ways of being 1 letter. (assuming no difference is there between upper case letter and lower case letter).

And that 1 remaining character get 10 ways of being a special character.

So, by product rule, this choice (without ordering) can be done in:

$26 \times 26 \times 26 \times 10 = 175760$ ways.

Now, the password may look like one of those:

1. L, L, L, S
2. L, L, S, L
3. L, S, L, L
4. S, L, L, L

where S shows presence of special character and L shows presence of letter.

Those 175760 ways are available for each of those four ways.

Thus, resultant number of ways this can be done is:

$175760 \times 4 = 703040$

Thus, the number of possibilities for constructing the 4 characters long password with specified conditions is given by: Option C: 703,040

Learn more about rule of product here:

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