How many different 3letter initials can people have How many

How many different 3-letter initials can people have? How many bit strings are there of length 8 or less, not counting the empty string? How many positive integers between 1000 and 9999 inclusive... are divisible by nine? are even? Have distinct digits? Are not divisible by 3? Are divisible by either 5 or 7? are not divisible by either 5 or 7? are divisible by 5 but not by 7? are divisible by 5 and 7?

Solution

1) To check how many different 3 letter initials people can have :

We have 26 alphabets in english letters, so people can have 26 preferences or 26 at second place or in third initial preference. So chances are 26*26*26 which is 17576 of three- lettter initails.

2) We have a formula for this which is 2^n , there are 2^n bits tring where n specifies the length Hence by formula we have 2^8 i.e 256.

3) a) As there are 9*10*10*10 numbers between 1000 and 9999 and exactly every 9th one is divisble by 9 so answer is 9000/9.

b) As there are 9*10*10*10 number between 1000 and 9999 and exactly every 2nd one is even so answer is 9000/2

c) We have four digits number and for the first one we have 9 distinct choices for second one 8 third one 7 and fourth one 6 hence answer is 9*8*7*6

d) We can easily count which are divisble by 3 and which is 9000/3 and now to take the one which are not divisble by 3 is the complement of divisble by 3 count hence 9000-9000/3

e) The numbers which are divisble by 5 are 9000/5=1800 and 9000/7=1285.81 are divisible by 7 But here we need to analyse to determinem if it is actually 1285 or 1286. I.e. in every group 1000-1006, 1007-1013, etc there is exactly one number which is divisible by 7. The last group starts with 9995=1000+7*1285 but of course gets cut off at 9999. So it just depends on where the (0 mod 7) element sits. It turns out it is 9996 hence that group does add one more multiple of 7.
Now what is the trouble with just taking 1800+1286? Well numbers like 35, 70, ... get counted twice. More generally |A union B| is equal to |A| + |B| - |A intersect B|. A intersect B here is the multiples of 35. By the last part of this question, this quantity is 257. So the answer to this part is 1800 + 1286 - 257.

f) To answer this one we will use the reference of e) part where we had find out number divisble by either 5 or 7 and complement of that will be numbers not divisble by 5 or 7 which is  (9000 - (1800 + 1286 - 257) )

g) This is the same as the set of those that are divisible by 5 and not 35, so we just take the number divisible by 35 away from the number that are divisible 5 (Set theoretically we can take the number that are divisible by 7 away from those that are divisible by 5 but just like with union this quantity is not the same as the difference of their numbers because not every multiple of 7 is in the set of multiples of 5). Answer is 1800 - 257.

h) Here it ask divisble by 5 and 7 which means divisble by multiplcation of both which is 35.Now which is approx 9000/35.  29*35 = 1015 and 285*35 = 9975. Every number from 29 to 285 inclusive multiplied by 35 gives one of our desired numbers. There are 285 - 29 + 1 = 257.