Question

(a) \\(\frac{6\cdot5\cdot4\cdot3\cdot2\cdot1}{5\cdot4\cdot3\cdot2\cdot1} = \\) type your answer...
(b) \\(\frac{8\cdot7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1}{6\cdot5\cdot4\cdot3\cdot2\cdot1} = \\) type your answer...
(c) \\(3! = \\) type your answer...
(d) \\(6! = \\) type your answer...
(e) \\(\frac{4!}{3!} = \\) type your answer...
(f) \\(\frac{6!}{4!} = \\) type your answer...
(g) \\(\frac{10!}{99!} = \\) type your answer... you will have to compute this one by hand. lots of numbers will cancel!
(h) \\(_5p_2 = \\) type your answer...
(i) \\(_5p_5 = \\) type your answer...
(j) \\(p(7, 3) = \\) type your answer...
(k) \\(_6c_4 = \\) type your answer...
(l) \\(c(8, 3) = \\) type your answer...

Explanation:

Part (a)

Step 1: Identify the factorials

The numerator \(6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1\) is \(6!\) and the denominator \(5 \cdot 4 \cdot 3 \cdot 2 \cdot 1\) is \(5!\).

Step 2: Simplify the fraction

We know that \(n! = n\times(n - 1)!\), so \(\frac{6!}{5!}=\frac{6\times5!}{5!}\). The \(5!\) terms cancel out, leaving \(6\).

Part (b)

Step 1: Identify the factorials

The numerator \(8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1\) is \(8!\) and the denominator \(6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1\) is \(6!\).

Step 2: Simplify the fraction

Using the property \(n! = n\times(n - 1)\times\cdots\times k!\) for \(n>k\), we have \(\frac{8!}{6!}=\frac{8\times7\times6!}{6!}\). The \(6!\) terms cancel out, and \(8\times7 = 56\).

Part (c)

Step 1: Recall the definition of factorial

The factorial of a non - negative integer \(n\) is defined as \(n!=n\times(n - 1)\times\cdots\times1\) for \(n\geq1\) and \(0! = 1\). For \(n = 3\), \(3!=3\times2\times1\).

Step 2: Calculate the product

\(3\times2\times1=6\).

Part (d)

Step 1: Recall the factorial formula

For \(n = 6\), \(6!=6\times5\times4\times3\times2\times1\).

Step 2: Calculate the product

\(6\times5 = 30\), \(30\times4=120\), \(120\times3 = 360\), \(360\times2=720\), \(720\times1 = 720\).

Part (e)

Step 1: Identify the factorials

We have \(\frac{4!}{3!}\), and \(4!=4\times3!\).

Step 2: Simplify the fraction

\(\frac{4\times3!}{3!}\), the \(3!\) terms cancel out, leaving \(4\).

Part (f)

Step 1: Identify the factorials

\(\frac{6!}{4!}\), and \(6!=6\times5\times4!\).

Step 2: Simplify the fraction

\(\frac{6\times5\times4!}{4!}\), the \(4!\) terms cancel out. Then \(6\times5 = 30\).

Part (g)

Step 1: Identify the factorials

\(\frac{101!}{99!}\), and \(101!=101\times100\times99!\).

Step 2: Simplify the fraction

\(\frac{101\times100\times99!}{99!}\), the \(99!\) terms cancel out. Then \(101\times100=10100\).

Part (h)

Step 1: Recall the permutation formula

The permutation formula is \(_{n}P_{r}=\frac{n!}{(n - r)!}\), where \(n = 5\) and \(r = 2\).

Step 2: Substitute the values

\(_{5}P_{2}=\frac{5!}{(5 - 2)!}=\frac{5!}{3!}\). Since \(5!=5\times4\times3!\), then \(\frac{5\times4\times3!}{3!}=5\times4 = 20\).

Part (i)

Step 1: Recall the permutation formula

For \(_{n}P_{n}\), using the formula \(_{n}P_{r}=\frac{n!}{(n - r)!}\) with \(r=n\), we get \(_{n}P_{n}=\frac{n!}{(n - n)!}=\frac{n!}{0!}\). Since \(0! = 1\), \(_{n}P_{n}=n!\). For \(n = 5\), \(_{5}P_{5}=5!\).

Step 2: Calculate \(5!\)

\(5!=5\times4\times3\times2\times1 = 120\).

Part (j)

Step 1: Recall the permutation formula

The formula for \(P(n,r)=\frac{n!}{(n - r)!}\), where \(n = 7\) and \(r = 3\).

Step 2: Substitute the values

\(P(7,3)=\frac{7!}{(7 - 3)!}=\frac{7!}{4!}\). Since \(7!=7\times6\times5\times4!\), then \(\frac{7\times6\times5\times4!}{4!}=7\times6\times5\). \(7\times6 = 42\), \(42\times5=210\).

Part (k)

Step 1: Recall the combination formula

The combination formula is \(_{n}C_{r}=\frac{n!}{r!(n - r)!}\), where \(n = 6\) and \(r = 2\).

Step 2: Substitute the values

\(_{6}C_{2}=\frac{6!}{2!(6 - 2)!}=\frac{6!}{2!4!}\). Since \(6!=6\times5\times4!\), we have \(\frac{6\times5\times4!}{2\times1\times4!}\). The \(4!\) terms cancel out, and \(\frac{6\times5}{2\times1}=\frac{30}{2}=15\).

Part (l)

Step 1: Recall the combination formula

The formula for \(C(n,r)=\frac{n!}{r!(n - r)!}\), where \(n = 8\) and \(r = 3\).

Step 2: Substitute the values

\(C(8,3)=\frac{8!}{3!(8 - 3)!}=\frac{8!}{3!5!}\). Since \(8!=8\times7\times6\times5!\), we have \(\frac{8\times7\times6\times5!}{3\times2\times1\times5!}\). The \(5!\) terms cancel out, and \(\frac{8\times7\times6}{3\times2\times1}=\frac{336}{6}=56\).

Answers:

(a) \(\boldsymbol{6}\)

(b) \(\boldsymbol{56}\)

(c) \(\boldsymbol{6}\)

(d) \(\boldsymbol{720}\)

(e) \(\boldsymbol{4}\)

(f) \(\boldsymbol{30}\)

(g) \(\boldsymbol{10100}\)

(h) \(\boldsymbol{20}\)

(i) \(\boldsymbol{120}\)

(j) \(\boldsymbol{210}\)

(k) \(\boldsymbol{15}\)

(l) \(\boldsymbol{56}\)

Answer:

s:
(a) \(\boldsymbol{6}\)

(b) \(\boldsymbol{56}\)

(c) \(\boldsymbol{6}\)

(d) \(\boldsymbol{720}\)

(e) \(\boldsymbol{4}\)

(f) \(\boldsymbol{30}\)

(g) \(\boldsymbol{10100}\)

(h) \(\boldsymbol{20}\)

(i) \(\boldsymbol{120}\)

(j) \(\boldsymbol{210}\)

(k) \(\boldsymbol{15}\)

(l) \(\boldsymbol{56}\)