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

Step-by-step explanation:

Given 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.

Thus we find that 2 letters can be selected in

Since for numbers repitition is not allowed, 4 digits can be arranged in 9P4 ways

= 9*8*7*6

Total no of ways for I type=

For II type numbers are 2 and letters are 4.

So no of ways =

total referral codes possible =

mr. flanders’ class sold 47 doughnuts and   mr. rodriquez’s class sold 38 cartons of milk

step-by-step explanation:

let x be the number of doughnuts sold by mr. flanders’ class and y be the number of cartons sold by mr. rodriquez’s class.

1. together, the classes sold 85 items, then

x+y=85.

2. both classes earn earned $104.49 for their school. if doughnuts were sold for $1.35 each, then x doughnuts costed $1.35x. if cartons of milk were sold for $1.08 each, then y cartons costed $1.08y. hence,

1.35x+1.08y=104.49.

3. solve the system of two equations:

£56

step-by-step explanation: