Question

We usually write numbers in decimal form (or base 10), meaning numbers are composed using 10 different “digits” {0,1,…,9}.{0,1,…,9}. Sometimes though it is useful to write numbers hexadecimal or base 16. Now there are 16 distinct digits that can be used to form numbers: {0,1,…,9,A,B,C,D,E,F}.{0,1,…,9,A,B,C,D,E,F}. So for example, a 3 digit hexadecimal number might be 2B8.

How many 3-digit hexadecimals are there in which the first digit is E or F?

How many 4-digit hexadecimals start with a letter (A-F) and end with a numeral (0-9)?

How many 4-digit hexadecimals start with a letter (A-F) and end with a numeral (0-9)?

In a recent survey, 43 students reported whether they liked their potatoes Mashed, French-fried, or Twice-baked. 24 liked them mashed, 23 liked French fries, and 12 liked twice baked potatoes. Additionally, 11 students liked both mashed and fried potatoes, 9 liked French fries and twice baked potatoes, 10 liked mashed and baked, and 2 liked all three styles. How many students hate potatoes? Explain why your answer is correct.

How many positive integers less than 975 are multiples of 8, 7, or 9? Use the Principle of Inclusion/Exclusion.

We want to build 5 letter “words” using only the first n=12n=12 letters of the alphabet. For example, if n=5n=5 we can use the first 5 letters, {a,b,c,d,e}{a,b,c,d,e} (Recall, words are just strings of letters, not necessarily actual English words.)

1. How many of these words are there total?
2. How many of these words contain no repeated letters?
3. How many of these words start with the sub-word “ade”?
4. How many of these words either start with “ade” or end with “be” or both?
5. How many of the words containing no repeats also do not contain the sub-word “bed”?

Answer 1

1.

a) First digit is E or F. So two options. Second digit is any one out of 16 options and third one is also so. This logic appears as there is no bound about repetition of digits.

So, number of 3 digited hexadecimals with the given condition is = 2*16*16 = 512.

b) First digit is any one of A-F and so 6 options, fourth one is any one of 0-9 and so 10 options and remaining two places each with 16 options.

So, number of 4 digited hexadecimals with the given conditions is = 6*16*16*10 = 15360.

3. n(divisible by 8,7 or 9) = n(divisible by 8) + n(divisible by 7) + n(divisible by 9) - n(divisible by 8*7) - n(divisible by 7*9) - n(divisible by 9*8) + n(divisible by 8*7*9)

=121+139+108-17-15-13+1 = 324.

4.

a) Total words = 12^5 = 248832 ;as repetition is allowed.

b) Words without repeated letters = 12*11*10*9*8 = 95040    ;as first place holds any one of 12, second place holds any one of remaining 11, third place holds any one of remaining 10, fourth place holds any one of remaining 9 and fifth place holds any one of remaining 8.

c) Number of words start with "ade" = 12*12=144    ; as no restriction about repetition.

d) Number of such words = n(words start with "ade") + n(words end with "be") -

   n(words start with "ade" and end with "be")

= 12*12 + 12*12*12 -1 = 144+1728-1 = 1871

e) Number of such words = n(words without repetitions of letters) - n(words without repetition of letters that contain sub-word "bed")

   = 95040 - 3*(9*8*) = 94824 ;as "bed" takes one of the 3 places while remaining 2 places are filled by remaining 9 and 8 letters respectively.