Permutations And Combinations

To determine how many six-digit numbers can be formed using the digits 0, 1, 2, 3, 4, 5, 6, and 7, we need to consider the number of choices for each digit position.

For the first digit position (leftmost), we have 7 choices (excluding 0 since a six-digit number cannot start with 0).

For the remaining five digit positions (from left to right), we have 8 choices (including 0) since all digits are available.

Therefore, the total number of six-digit numbers that can be formed is obtained by multiplying the number of choices for each digit position:

Total number of six-digit numbers = 7 * 8 * 8 * 8 * 8 * 8 = 7 * 8^5 = 7 * 32768 = 229,376

Hence, there are 229,376 six-digit numbers that can be formed using the digits 0, 1, 2, 3, 4, 5, 6, and 7.

Counting 3-term Geometric Progressions in . We want to count the number of ways to pick three \emph{distinct} numbers such that form a Geometric Progression (GP). In particular, this means there exists a common ratio (possibly rational) so that Equivalently, the GP condition on distinct positive integers can be written as Hence the counting problem reduces to finding all integer triples with satisfying . \medskip

Step-by-step Enumeration Strategy.
(A) Fix an integer in the range .
(B) Look at the divisors of . For each divisor of , set
(C) We then check the conditions:
(D) Count every such valid triple .

Why this works. If , then necessarily is a divisor of , and . The condition ensures the three terms are distinct and in ascending order. Finally, we require . A short computer‐based enumeration reveals there are exactly such triples. Hence, the number of ways of selecting three numbers in GP from is

The number of subsets of a set with 8 elements is . We are asked to find the number of subsets that contain at least 6 elements.

Step 1: Use the binomial coefficient to calculate the number of subsets with exactly 6, 7, and 8 elements:

Step 2: The total number of subsets with at least 6 elements is 37.