- Compare what the arrangement looks like before and after exchanging the two identical symbols. - Ask how many orders the two copies of A would have if they were temporarily labeled.

1. No. The two A's are identical, so exchanging them does not change the visible arrangement. 2. Treating them as distinct would count each visible arrangement \(2!\) times.

No; the overcounting factor is \(2!\).

- Identify every symbol that occurs more than once. - Each repeated group creates its own overcounting factor.

1. There are two identical M's and two identical A's. 2. Correct for both repeated groups, so the appropriate expression is \(\frac{4!}{2!\,2!}\).

\(\frac{4!}{2!\,2!}\)

- Begin with one fixed first letter and list the different ways to complete the row. - Repeat with the other first letter. - Check whether swapping the two identical A's creates a new arrangement.

1. Start with AA, which gives AABB, or end with AA, which gives BBAA. 2. The remaining arrangements are ABAB, ABBA, BAAB, and BABA. 3. There are \(6\) distinct arrangements. Swapping the two A's or the two B's cannot create another one.

AABB, ABAB, ABBA, BAAB, BABA, BBAA; \(6\) arrangements.

- What information defines a tile color pattern? - What information defines where named students sit?

1. The red tiles look the same in the color pattern, so swapping them does not change the pattern. 2. Noor and Felix are different people, so switching their seats does change which person occupies each position.

Red tiles: no new pattern. Noor and Felix: a different seating arrangement, because their positions are distinguishable.

- Which square makes one pattern different from another? - After you place the orange square, can moving the blue squares change the pattern?

1. The orange square can be in position \(1\), \(2\), \(3\), or \(4\). 2. Once its position is chosen, all the other squares are blue, so there are no more choices. 3. There are \(4\) patterns.

\(4\) patterns, one for each possible position of the orange square.

- For part a, count the permutations of three distinct objects. - In part b, treat the first red bead as fixed. - How many distinct orders can be made from one red bead and two identical blue beads?

a) The three distinct beads can be arranged in \(3!=6\) orders. b) Fix one red bead in the first position. The remaining beads are one red and two identical blue beads, so the number of distinct arrangements is \(\frac{3!}{2!}=3\).

a) \(6\) arrangements b) \(3\) arrangements

- How many permutations are possible when all three digits are distinct? - For part b, list the codes in increasing order. - Why do two identical digits reduce the number of distinct arrangements?

a) Three distinct digits can be arranged in \(3!=6\) orders. b) For the multiset \(\{1,1,2\}\), divide by the \(2!\) arrangements of the identical \(1\)s: \(\frac{3!}{2!}=3\). The codes are \(112\), \(121\), and \(211\).

a) \(6\) codes b) \(3\) codes: \(112,121,211\)

- Count the total number of letters and identify which letters repeat. - Swapping two identical copies does not create a new arrangement. - Account separately for each group of repeated letters.

1. LEVEL has \(5\) letters, with \(2\) identical L's and \(2\) identical E's. 2. Start with \(5!\) arrangements and divide by \(2!\) for the repeated L's and by \(2!\) for the repeated E's: \(\frac{5!}{2!\,2!}=30\).

\(30\) distinct arrangements

- Fix the required first tile before arranging the others. - Which letters are still repeated? - How many labeled orders represent each visible arrangement?

1. Fix a P in the first position. The remaining tiles are M, M, N, N, and P. 2. If all five remaining tiles were different, there would be \(5!\) orders. Swapping the two M's or the two N's does not change an arrangement. 3. Divide by the repeats: \(\frac{5!}{2!2!}=30\) distinct arrangements.

\(30\) arrangements.

- Try replacing the labels \(E_1,E_2,E_3\) with ordinary E's. Which arrangements become identical? - How many ways can three labels be exchanged while leaving the visible word unchanged? - What correction should be made to the labeled count?

1. In any fixed visible arrangement, the three E tiles can exchange labels in \(3!\) ways without changing the letters shown. 2. The \(5!\) labeled arrangements therefore count every visible arrangement \(3!\) times. 3. Divide to obtain \(\frac{5!}{3!}=20\) distinct arrangements.

\(20\) arrangements. Each visible arrangement is counted \(3!=6\) times when the E tiles are labeled.

- Follow a branch from the start all the way to its last tile. - Which differences between paths would disappear if the small labels were removed? - Compare two paths that begin with different A tiles.

1. The tree shows six complete paths, one for each order of the three temporarily distinguishable tiles. 2. Erasing the small labels makes the two A tiles identical. The visible rows are AAB, ABA, and BAA, so there are three different rows. 3. For example, \(A_1\)–\(A_2\)–B and \(A_2\)–\(A_1\)–B are different paths but both become AAB when the small labels are erased. Every visible row is counted twice in the tree.

a) \(6\) paths b) \(3\) different rows: AAB, ABA, BAA c) For example, \(A_1\)–\(A_2\)–B and \(A_2\)–\(A_1\)–B both become AAB.

- After fixing the first binder, identify the binders that remain. - Does switching two identical binders create a new arrangement? - Compare a multiset with two identical objects to three distinct objects.

a) Fix the red binder at the far left. The remaining binders are blue, blue, and green, giving \(\frac{3!}{2!}=3\) distinct arrangements. b) Fix a blue binder at the far left. The remaining binders are blue, red, and green, all distinguishable, so there are \(3!=6\) arrangements. c) In part a, interchanging the two identical blue binders does not create a new arrangement. In part b, all three remaining binders are different.

a) \(3\) arrangements b) \(6\) arrangements c) In part a, two remaining binders are identical; in part b, all three remaining binders are distinct.

- Begin with the possible hundreds digits \(2,5,\) and \(8\). - Remember that the two \(8\) cards are identical. - After forming a three-digit number, how many cards remain unused? - For part b, compare directly with part a or use the repeated-object permutation count.

a) The distinct three-digit numbers are \(258,285,288,528,582,588,825,828,852,858,882,\) and \(885\), for a total of \(12\). b) The number of distinct permutations of all four cards is \(\frac{4!}{2!}=12\). Each distinct three-digit arrangement leaves exactly one unused card, which can be appended to form a corresponding four-digit arrangement, giving a one-to-one correspondence.

a) \(12\) three-digit numbers b) \(12\) four-digit numbers. Each three-digit arrangement determines exactly one four-digit arrangement by appending the unused card.

- Determine how many arrangements a triple such as \((1,2,6)\) has. - Compare triples with three distinct entries, two equal entries, and three equal entries. - The equally likely elementary outcomes identify the result on each of the three distinguishable dice.

1. For sum \(9\), the unordered triples are \((1,2,6)\), \((1,3,5)\), \((1,4,4)\), \((2,2,5)\), \((2,3,4)\), and \((3,3,3)\). For sum \(10\), they are \((1,3,6)\), \((1,4,5)\), \((2,2,6)\), \((2,3,5)\), \((2,4,4)\), and \((3,3,4)\). Each sum has \(6\) unordered triples. 2. There are \(216\) equally likely ordered outcomes. For sum \(9\), the multiplicities give \(3\cdot6+2\cdot3+1=25\) outcomes, so \(P(X=9)=\frac{25}{216}\approx0.1157\). For sum \(10\), the multiplicities give \(3\cdot6+3\cdot3=27\) outcomes, so \(P(X=10)=\frac{27}{216}=0.125\). 3. Unordered triples are not equally likely: a triple with three distinct entries represents \(6\) ordered outcomes, one with two equal entries represents \(3\), and one with three equal entries represents \(1\). Thus the unordered count alone ignores different arrangement multiplicities.

1. Each sum has \(6\) unordered triples. 2. \(P(X=9)=\frac{25}{216}\approx0.1157\) and \(P(X=10)=\frac{27}{216}=0.125\) 3. The unordered triples represent different numbers of equally likely ordered outcomes, depending on whether entries repeat.

- The four groups have fixed sizes, but the students assigned to them can change. - How often does each group letter appear in the six-position string? - Would exchanging two copies of the same group letter create a different assignment?

1. One possible six-letter assignment in the given student order is AABCDD. Each full assignment has two A's, one B, one C, and two D's. 2. Labeling the two A's and two D's separately would give \(6!\) arrangements. Swapping the two A's or the two D's leaves the group assignments unchanged. 3. Each different group assignment is counted \(2!\cdot2!\) times, so there are \(\frac{6!}{2!\cdot2!}=180\) assignments.

One possible assignment is AABCDD in the given student order. There are \(\frac{6!}{2!\cdot2!}=180\) different assignments; swapping identical group letters does not change an assignment.

- For the total count, account for every repeated symbol group. - In part b, impose the required first two symbols before counting what remains. - Compare the favorable distinct-string count with the total distinct-string count.

a) The total number of distinct strings is \(\frac{8!}{3!\,3!\,2!}=560\). b) If the first two positions are A's, the remaining six positions contain one A, three B's, and two C's. The number of favorable strings is \(\frac{6!}{1!\,3!\,2!}=60\). Therefore, the probability is \(\frac{60}{560}=\frac{3}{28}\).

a) \(560\) distinct strings b) \(\frac{3}{28}\approx0.1071\)

- Start from the number of arrangements of six distinct positions, then account for swaps among identical A's. - Express the arrangement count in terms of the unknown multiplicity. - Look for a factorial value that matches the resulting denominator.

1. If \(m\) copies are identical, the number of distinct arrangements is \(\frac{6!}{m!}\). 2. Set \(\frac{6!}{m!}=120\). Since \(6!=720\), this gives \(m!=6\). 3. Because \(3!=6\), \(m=3\).

\(m=3\)

- Compare the adjacency condition with the unrestricted arrangements. - Look for a way to enforce adjacency before counting rather than correcting afterward. - After the adjacency condition is built into the objects being arranged, identify any letters that still repeat.

1. Dividing the unrestricted count by \(2\) assumes that exactly half of all distinct arrangements have adjacent O’s, which is not justified. 2. Enforce the adjacency condition before counting by treating the two O’s together as one unit. The objects to arrange are the OO unit, B, A, L, L, and N: \(6\) objects with two identical L’s. 3. The correct count is \(\frac{6!}{2!}=360\).

Priya’s division by \(2\) is unjustified. The correct number of arrangements is \(360\).

- Try a smaller version with two A's. Which rows keep them apart? - What could you count first so that the remaining restriction becomes easier to handle? - Does the condition change how many positions are available for the A's?

1. Arrange the non-A letters B, B, C in \(\frac{3!}{2!}=3\) visible orders. 2. Every such order creates four gaps: before, between, and after these letters. To keep the A's apart, place each of the three identical A's in a different gap. 3. Choose three of the four gaps, giving \(\binom{4}{3}=4\) placements for every B, B, C order. 4. The total is \(3\cdot4=12\) distinct arrangements.

\(12\) arrangements.