Question

How many positive integers less than 1150 are multiples of 2,5 , or 9 ? Use the Principle of Inclusion/Exclusion. We want to build 5 letter "words" using only the first n=10 letters of the alphabet. For example, if n=5 we can use the first 5 letters, {a,b,c,d,e} (Recall, words are just strings of letters, not necessarily actual English words.) a. How many of these words are there total? b. How many of these words contain no repeated letters? c. How many of these words start with the sub-word "ade"? d. How many of these words either start with "ade" or end with "be" or both? e. How many of the words containing no repeats also do not contain the sub-word "bed"?

Answer 1

The number of positive integers less than 1150 that are multiples of 2, 5, or 9 is 740. There are a total of 100,000 possible 5-letter 'words' from the first ten letters of the alphabet, 30,240 without repeated letters, 64 starting with 'ade', 1,063 starting with 'ade' or ending with 'be', and a calculated number excluding the sequence 'bed'.

Step-by-step explanation:

To find how many positive integers less than 1150 are multiples of 2, 5, or 9, we apply the Principle of Inclusion/Exclusion. First, count the multiples of each number. For 2, there are ⅔(1150) = 574 multiples. For 5, ⅔(1150) = 229, and for 9, ⅔(1150) = 127. Next, find the least common multiples (LCM) for each pair to count the numbers that are multiples of both. LCM of 2 and 5 is 10, so ⅔(1150) = 114 multiples. LCM of 2 and 9 is 18, so ⅔(1150) = 63. LCM of 5 and 9 is 45, so ⅔(1150) = 25. Lastly, for numbers that are multiples of 2, 5, and 9, LCM is 90, ⅔(1150) = 12. Now apply inclusion-exclusion: total multiples = 574 + 229 + 127 - (114 + 63 + 25) + 12 = 740.

For the 5-letter 'words' using the first n=10 letters of the alphabet:

• a. The total number of such words is 10⁵ = 100,000.
• b. Words with no repeated letters can be calculated using permutation, 10P5 = 30,240.
• c. Words starting with 'ade' have 2 free spots left, so 8² = 64 possibilities.
• d. Words starting with 'ade' are the 64 counted in c. For words ending in 'be', there are 10³ = 1,000 options. However, the overlap of words starting with 'ade' and ending with 'be' must be subtracted. There's only 1 such word, so the total is 64 + 1,000 - 1 = 1,063.
• e. For no repeats and excluding 'bed', count all permutations without 'b', 'e', and 'd', which is 7P5. Then add back permutations where 'b', 'e', or 'd' occur but not in the sequence 'bed'.