1.How many positive three-digit integers can be made from the digits {3,4,5,6,7} if digits may be repeated? 5x4x3=60
2.In how many ways can all of the letters of the word HEXAGON be arranged if the arrangement must start with a consonant and end in a vowel? 4x3x1x2x3x4x5=1440
3.How many positive integers between 3000 and 4000 can be made with the digits {3,4,6.9} if repetition of digits is allowed? 3x1x2x3
4.How many 7-digit phone numbers can be made if the first three digits can only be 3's or 4's but there are no restrictions on the last four digits?(Don't know this one)
#1. 5 choices for each digit.
So, 5^4
#2. correct
#3. 1 * 4^3
#4. 2^3 * 10^4
when repetitions are allowed, you get more choices...
To find the number of positive three-digit integers that can be made from the digits {3,4,5,6,7} if digits may be repeated, you can use the principle of counting.
There are 5 choices for the first digit (any of the digits {3,4,5,6,7}), 5 choices for the second digit (any of the digits {3,4,5,6,7}), and 5 choices for the third digit (again, any of the digits {3,4,5,6,7}).
Therefore, the total number of possible three-digit integers is obtained by multiplying the number of choices for each digit:
5 x 5 x 5 = 125
So, there are 125 positive three-digit integers that can be made from the given digits when repetition is allowed.
Regarding the second question, to find the number of ways all the letters of the word HEXAGON can be arranged if the arrangement must start with a consonant and end in a vowel, you can again use the principle of counting.
The word HEXAGON has 7 letters, consisting of 2 vowels (E and O) and 5 consonants (H, X, G, N, and G).
To start with a consonant, there are 5 choices (H, X, G, N, or G). After selecting a consonant for the first letter, there are 6 letters left.
To end with a vowel, there are 2 choices (E or O). After selecting a vowel for the last letter, there are 5 letters remaining.
For the remaining 5 letters, there are 5! (5 factorial) ways to arrange them.
Therefore, the total number of arrangements is obtained by multiplying the number of choices for each position:
5 x 6 x 5! = 1440
So, there are 1440 ways to arrange the letters of the word HEXAGON if the arrangement must start with a consonant and end in a vowel.
Regarding the third question, to find the number of positive integers between 3000 and 4000 that can be made with the digits {3,4,6,9} with repetition of digits allowed, you can again use the principle of counting.
For the first digit, there are 4 choices (3, 4, 6, or 9). For the second digit, there is 1 choice (only one digit available since repetition is allowed).
For the third digit, there are 4 choices (again, any of the digits {3, 4, 6, 9}). For the fourth digit, there are 3 choices (any of the digits {3, 4, 6, 9} except the digit used in the third position).
Therefore, the total number of possible integers is obtained by multiplying the number of choices for each digit:
4 x 1 x 4 x 3 = 48
Therefore, there are 48 positive integers between 3000 and 4000 that can be made with the given digits if repetition is allowed.
Regarding the fourth question, to find the number of 7-digit phone numbers that can be made if the first three digits can only be 3's or 4's but there are no restrictions on the last four digits, you can again use the principle of counting.
For the first digit, there are only 2 choices (3 or 4). For the second and third digits, there are also only 2 choices (either 3 or 4).
For the remaining four digits (the last four digits), there are 10 choices for each digit (from 0 to 9).
Therefore, the total number of possible phone numbers is obtained by multiplying the number of choices for each digit:
2 x 2 x 2 x 10 x 10 x 10 x 10 = 160,000
Therefore, there are 160,000 different 7-digit phone numbers that can be made if the first three digits can only be 3's or 4's but there are no restrictions on the last four digits.