Permutations and Combinations: questions with answers

Equal combination counts from the same pool can have equal or complementary group sizes. Since the two lower indices differ, they must add to n.

Fixing 0 in the leading place uses up the 0, so the two other places share the remaining 8 digits. Giving each of those places all 9 digits again counts arrangements that repeat a digit.

Nothing has been used before the first place, so it has 9 choices. There are three places, and each used letter becomes unavailable.

The allowed flag counts run from 3 through 9, including both endpoints. Counting only 3 flags treats the minimum as an exact requirement.

The falling product counts ordered choices. Dividing by 3! removes the six orders of every three-object selection.

Without repetition each digit used is gone, so the three places have 9, one fewer, and one fewer again. Giving every place all 9 digits counts numbers that repeat a digit.

The denominator's number is the count of objects left over, so reading it straight off as the lower index is wrong every time -- 6 and 4 are never equal in this domain. A lower index one larger belongs to the product carried one factor too far.

The number of surviving factors is the gap between the two numbers. Running the product down to 6 itself includes one extra factor.

A signal cannot use both exactly two and exactly three flags. Add the two alternative counts; multiplying would count pairs of signals.

Pooling everyone and choosing four gives the largest of the three numbers in every binding, and adding the two counts gives the smallest, so the three separate cleanly. Pooling loses the requirement of two from EACH group; adding answers a question with 'or' in it where this one has 'and'.

First count the 7 letter copies as distinct, giving 7!. Rearranging the 4 identical copies does not change the word, so each distinct word is counted 4! times. Divide: $7!/ 4!=210$.

Each occupied position supplies one factor in the falling product. There are three positions, so the product contains 3 factors. The number of available letters determines the first factor, not how many factors there are.

Cancelling is division, so it takes the SAME factor off both sides; the little numbers 4 and 2 are labels on the symbols and cannot be cancelled against each other. One wrong line stops the cancelling a factor early and one changes the signs, which are the two slips the chapter's Example 12 has to be worked carefully to avoid.

There are 5 gaps between the 6 girls, plus one place before the first girl and one after the last. A boy can sit in either end position without being next to another boy. With at most one boy in each gap, the total is $5 + 2 = 7$ gaps.

Seven points give 21 chords, which is the chapter's own worked instance; Exercise 6.4 Question 3 runs the same count over 21 points and answers 210. The three offered numbers -- the pair count, its double and twice the number of points -- are distinct for every point count here, so no two routes ever agree.