Question

1- Say 5 students forget to schedule meetings with their Math teacher for HW0, so they have to wait outside the office just in case their instructor finishes scheduled meetings early enough to squeeze them in. Say there is enough extra time for the instructor to meet 3 of these 5 students.

How many orders are there for how 3 of the waiting students can meet with their instructor?

2-Say that a referral code is 2 letters (A-Z, no lowercases) followed by 4 digits (0-9), or it is 2 digits (0-9) followed by four letters (A-Z, no lowercases). Digits are not allowed to repeat but letters can repeat. How many referral codes can you make?

Answer 1

Answer1:

Total student count , n = 5

count of students who can meet with instructor , r = 3

The different order in which the 3 students from total 5 students can meet with instructor ,

Note: We are considering Permutations here , as Order of students meeting is important.

So total orders , 3 students can meet with instructor = 60

Answer 2:

It is given that , Referral code can be of 2 Capital letters followed by 4 digits or 2 digits followed by 4 Capital letters , where digits are not allowed to repeat , and letters can repeat.

Total number of possible referral code = Possible count of ( 2 Capital letters followed by 4 digits) + Possible count of (2 digits followed by 4 Capital letters)

Total Capital letters count = 26 , ( A... Z) , Total digits count = 10 , ( 1,2...... 9)

We know that Capital letters can repeat , and digits can't be repeated in a referral code. So

Possible count of ( 2 Capital letters followed by 4 digits) = 26 * 26 * 10 * 9 * 8 * 7 = 34,07,040

Possible count of ( 2 digits followed by 4 letters) = 10 * 9 * 26 * 26 * 26 * 26 = 4,11,27,840

Total number of referral code possible = 34,07,040 + 4,11,27,840 = 4,45,34,880