Free math worksheets

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

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

- 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

- 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 table records category counts but not respondent identities. - Assign labels to cells in stages. - Equivalent assignments can be counted with a multinomial coefficient.

1. Choose which \(2\) respondents enter the first two-count cell: \(\binom{6}{2}\). 2. Choose \(1\) of the remaining \(4\) for the first one-count cell, then \(1\) of the remaining \(3\) for the second one-count cell. 3. The last \(2\) respondents enter the final cell. 4. The count is \(\binom{6}{2}\binom{4}{1}\binom{3}{1}=\frac{6!}{2!\,1!\,1!\,2!}=180\).

\(180\) labeled assignments.

- 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) \(\frac{3}{28}\approx0.1071\)

- 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 \(\frac{6!}{2!}=360\).

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