Free Math Worksheets

- How is an entry formed from the two entries above it? - Compare the left and right sides of a row. - What pattern appears at the two ends of every row?

1. Pascal's rule gives \(\binom{14}{7}=\binom{13}{6}+\binom{13}{7}=1716+1716=3432\). 2. The symmetry identity is \(\binom{n}{k}=\binom{n}{n-k}\). Since \(13-6=7\), \(\binom{13}{6}=\binom{13}{7}\). 3. Every row begins and ends with \(1\), so \(\binom{15}{0}=1\) and \(\binom{15}{15}=1\).

a) \(\binom{14}{7}=3432\) b) Since \(13-6=7\), symmetry gives \(\binom{13}{6}=\binom{13}{7}\). c) \(\binom{15}{0}=1\), \(\binom{15}{15}=1\)

- Recall the entries and indexing of row \(n=7\). - Which two values of \(k\) are symmetric in a row? - Replace a large \(k\) with \(n-k\) before calculating. - Compare the number of factors needed for \(\binom{10}{8}\) and \(\binom{10}{2}\).

1. Row \(n=7\) contains \(1,7,21,35,35,21,7,1\). Thus \(35=\binom{7}{3}=\binom{7}{4}\). 2. By symmetry, \(\binom{10}{8}=\binom{10}{2}\). 3. \(\binom{10}{2}=\frac{10\cdot 9}{2}=45\). Using \(k=2\) requires fewer factors than calculating directly with \(k=8\).

a) \(\binom{7}{3}\) and \(\binom{7}{4}\) b) \(\binom{10}{8}=\binom{10}{2}=45\)

- Match each entry in row \(n=5\) with its number of successes, beginning with \(0\). - “More than \(3\)” includes \(4\) and \(5\). - With equally likely paths, divide favorable paths by all paths.

1. The entry for exactly \(3\) successes is \(\binom{5}{3}=10\). 2. More than \(3\) successes means \(4\) or \(5\) successes, so the number of paths is \(\binom{5}{4}+\binom{5}{5}=5+1=6\). 3. Exactly one success has \(\binom{5}{1}=5\) equally likely paths out of \(32\), so the probability is \(\frac{5}{32}=0.15625\).

a) \(10\) b) \(6\) c) \(\frac{5}{32}=0.15625\)

- Compare the coefficients with the expansion of \((a + b)^3\). - A power of a linear factor equals zero only when the linear factor equals zero. - The exponent on the factor gives the multiplicity.

1. The identity \((a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3\) with \(a = x\) and \(b = 1\) gives \(f(x) = (x + 1)^3\). 2. Set \((x + 1)^3 = 0\). Then \(x + 1 = 0\), so \(x = -1\). 3. Because the factor \(x + 1\) occurs to the third power, the zero has multiplicity 3.

1. \(f(x) = (x + 1)^3\) 2. \(x = -1\) with multiplicity 3

- Add adjacent entries to form the next row. - Use the recurrence formula or the definition of a binomial coefficient. - Add the entries of row \(n=5\) and compare the result with \(2^5\).

1. Add adjacent entries from the given row: the next row begins \(1\), \(1+9=10\), \(9+36=45\), \(36+84=120\), \(84+126=210\), and \(126+126=252\). 2. \(\binom{10}{4}=120\cdot\frac{10-3}{4}=120\cdot\frac{7}{4}=210\). 3. Row \(n=5\) is \(1,5,10,10,5,1\). Its sum is \(1+5+10+10+5+1=32=2^5\).

a) \(1,10,45,120,210,252\) b) \(\binom{10}{4}=210\) c) \(1+5+10+10+5+1=32=2^5\)

- Form the next Pascal's triangle row by adding adjacent entries. - Match each coefficient with its number of correct answers. - “At least \(5\)” includes \(5\) and \(6\). - How many total answer patterns are possible for \(6\) two-choice questions?

1. Row \(n=6\) is \(1,6,15,20,15,6,1\). 2. Exactly \(3\) correct answers can occur in \(\binom{6}{3}=20\) answer patterns. 3. At least \(5\) correct answers means \(5\) or \(6\). The number of favorable patterns is \(\binom{6}{5}+\binom{6}{6}=6+1=7\). There are \(2^6=64\) equally likely patterns, so the probability is \(\frac{7}{64}=0.109375\).

a) \(1,6,15,20,15,6,1\) b) \(20\) c) \(\frac{7}{64}=0.109375\)

- Reversing the order in an odd power changes the sign: compare \((a - b)^3\) and \((b - a)^3\). - Use cubic binomial identities or expand the factors directly. - Read multiplicity from the exponent on a linear factor. - A quadratic factor with no real zeros does not add real zeros or change the multiplicity of an existing linear factor.

1. At \(x = 3\), all five functions equal \(0\). At \(x = 0\), the values are \(f_1(0) = -27\), \(f_2(0) = -27\), \(f_3(0) = 27\), \(f_4(0) = -27\), and \(f_5(0) = -27\). 2. Function \(f_1\) has the zero \(x = 3\) with multiplicity 3. Function \(f_3(x) = -(x - 3)^3\) has the same zero and multiplicity. Expanding \(f_5\) gives \((x - 3)^3\), so \(f_5\) also has \(x = 3\) with multiplicity 3. 3. Factor the difference of cubes: \(f_2(x) = x^3 - 27 = (x - 3)(x^2 + 3x + 9)\). The quadratic factor has no real zeros, so \(f_2\) has \(x = 3\) with multiplicity 1. Function \(f_4\) is already the same product and has the same zero and multiplicity. 4. The binomial expansion gives \((x - 3)^3 = x^3 - 9x^2 + 27x - 27\), so \(f_1 = f_5\). The difference-of-cubes identity gives \(f_2 = f_4\). Function \(f_3 = -f_1\), so it is not identical to any other listed function.

a) At \(x = 3\), all five values are \(0\). At \(x = 0\): \(f_1 = -27\), \(f_2 = -27\), \(f_3 = 27\), \(f_4 = -27\), \(f_5 = -27\). b) \(f_1\), \(f_3\), and \(f_5\): \(x = 3\) with multiplicity 3 \(f_2\) and \(f_4\): \(x = 3\) with multiplicity 1 c) \(f_1 = f_5\) and \(f_2 = f_4\).

- List the sequences with exactly two makes and one miss. - Count how many such sequences there are. - Compare that count with the coefficients in the expansion of \((p+q)^3\).

1. The probability of a make is \(p=0.8\), and the probability of a miss is \(q=1-0.8=0.2\). 2. Exactly two makes occur in the sequences MMX, MXM, and XMM, where M represents a make and X represents a miss. 3. Each sequence has probability \(p^2q=(0.8)^2\cdot 0.2=0.128\). Therefore, \(P(\text{exactly two makes})=3p^2q=3\cdot(0.8)^2\cdot 0.2=0.384\). 4. The binomial expansion is \((p+q)^3=p^3+3p^2q+3pq^2+q^3\). The term \(3p^2q\) represents exactly two made shots and one miss.

a) \(0.384=38.4\%\) b) The term is \(3p^2q\).

- Record the letter sequence on each path. - Think about the possible position of the single \(b\). - Recall what \(\binom{n}{k}\) counts.

1. A three-stage tree with choices \(a\) and \(b\) has \(2^3=8\) paths. 2. A path producing \(a^2b\) must contain two \(a\) factors and one \(b\) factor. The paths are \((a, a, b)\), \((a, b, a)\), and \((b, a, a)\), so there are \(3\) paths. 3. The coefficient \(\binom{3}{2}=3\) counts the ways to choose the two of the three positions occupied by \(a\). This is exactly the number of paths producing \(a^2b\).

a) \(3\) paths b) \(a\to a\to b\), \(a\to b\to a\), and \(b\to a\to a\) c) \(\binom{3}{2}=3\) counts the ways to place two \(a\) factors among three positions.