Free Math Worksheets

- 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?

1. The three distinct beads can be arranged in \(3!=6\) orders. 2. In part b, fix one red bead in the first position. The remaining beads are one red and two identical blue beads. 3. The number of distinct arrangements of the remaining three beads is \(\frac{3!}{2!}=3\). They are RBB, BRB, and BBR.

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?

1. Three distinct digits can be arranged in \(3!=6\) orders: \(123,132,213,231,312,\) and \(321\). 2. For the multiset \(\{1,1,2\}\), divide by the \(2!\) arrangements of the identical \(1\)s: \(\frac{3!}{2!}=3\). 3. The three distinct codes are \(112,121,\) and \(211\).

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

- 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.

1. If the red binder is fixed at the far left, the remaining binders are blue, blue, and green. Their number of distinct arrangements is \(\frac{3!}{2!}=3\). 2. If a blue binder is fixed at the far left, the remaining binders are blue, red, and green. These three binders are distinguishable, so they can be arranged in \(3!=6\) ways. 3. Part a has fewer arrangements because 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, use the repeated-object permutation formula or compare directly with part a.

1. The distinct three-digit numbers are \(258,285,288,528,582,588,825,828,852,858,882,\) and \(885\), for a total of \(12\). 2. The number of distinct permutations of all four cards is \(\frac{4!}{2!}=12\), because the two \(8\)s are identical. 3. Each distinct three-digit arrangement leaves exactly one unused card, which must be placed at the end to form a corresponding four-digit arrangement. This gives a one-to-one correspondence between the two sets.

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 \(6^3=216\) equally likely ordered outcomes. For sum \(9\), three triples have all distinct entries and contribute \(3\cdot 6\) arrangements, two have one repeated entry and contribute \(2\cdot 3\), and \((3, 3, 3)\) contributes \(1\). Thus, there are \(3\cdot 6+2\cdot 3+1=25\) outcomes, so \(P(X=9)=\frac{25}{216}\approx 0.1157\). 3. For sum \(10\), three triples have all distinct entries and three have one repeated entry. Thus, there are \(3\cdot 6+3\cdot 3=27\) outcomes, so \(P(X=10)=\frac{27}{216}=0.125\). 4. Unordered triples are not equally likely: a triple with three distinct entries represents \(6\) ordered outcomes, a triple with two equal entries represents \(3\), and a triple with three equal entries represents \(1\). Therefore, simply counting unordered triples ignores their different numbers of arrangements.

1. Each sum has \(6\) unordered triples. 2. \(P(X=9)=\frac{25}{216}\approx 0.1157\) 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.