How many "words" can be written using exactly five a's and no more than three b's (and no other letters)?

This is a problem of Permutations. We have 3 cases depending on the number of B's. Since no more than three B's can be used we can use either one, two or three B's at a time.

Case 1: Five A's and One B

Total number of letters = 6

Total number of words possible =

Case 2: Five A's and Two B's

Total number of letters = 7

Total number of words possible =

Case 3: Five A's and Three B's

Total number of letters = 8

Total number of words possible =

Total number of possible words will be the sum of all three cases.

Therefore, the total number of words that can be written using exactly five A's and no more than three B's (and no other letters) are 6 + 21 + 56 = 83

84

Step-by-step explanation

Case 1: Five A's and One B

Total number of letters = 6

Total number of words possible = 6

Case 2: Five A's and Two B's

Total number of letters = 7

Total number of words possible = 21

Case 3: Five A's and Three B's

Total number of letters = 8

Total number of words possible = 56

Case 4: Five A's and no B's

Total number of letters = 5

Total number of words possible = 1

Total number of possible words will be the sum of all three cases.

Therefore, the total number of words that can be written using exactly five A's and no more than three B's (and no other letters) are 1+6 + 21 + 56 = 84

see explanation

step-by-step explanation:

since the sequence is geometric there is a common ratio r between consecutive terms, that is

r = 4 ÷ - 2 = - 2

r = - 8 ÷ 4 = - 2

r = 16 ÷ 8 = - 2

the next term - 13 must be incorrect as it should be - 32 for the sequence to be geometric.

the first 4 terms are therefore geometric with common ratio r = - 2

it is a. 4.2 because 7.6-3.4 gets the remaining length of 4.2. i might be wrong but i am pretty sure this is the answer.