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Q25E

Expert-verifiedFound in: Page 414

Book edition
7th

Author(s)
Kenneth H. Rosen

Pages
808 pages

ISBN
9780073383095

**One hundred tickets, numbered \(1,2,3, \ldots ,100\), are sold to \(100\) different people for a drawing. Four different prizes are awarded, including a grand prize (a trip to Tahiti). How many ways are there to award the prizes if**

**a) there are no restrictions?**

**b) the person holding ticket \(47\) wins the grand prize?**

**c) the person holding ticket \(47\) wins one of the prizes?**

**d) the person holding ticket \(47\) does not win a prize?**

**e) the people holding tickets \(19\) and \(47\) both win prizes?**

**f) the people holding tickets \(19\;,\;47\)and \(73\) all win prizes?**

**g) the people holding tickets \(19\;,\;47\;,\;73\) and \(97\) all win prizes?**

**h) none of the people holding tickets \(19\;,\;47\;,\;73\) and \(97\) wins a prize?**

**i) the grand prize winner is a person holding ticket \(19\;,\;47\;,\;73\) or \(97\)?**

**j) the people holding tickets 19 and 47 win prizes, but the people holding tickets \(73\) and \(97\) do not win prizes?**

a) The number of ways is \(94,109,400\).

b) The number of ways is \(941,094\).

c) The number of ways is \(3,764,376\).

d) The number of ways is \(90,345,024\).

e) The number of ways is \(114,072\).

f) The number of ways is \(2328\).

g) The number of ways is \(24\).

h) The number of ways is \(79,727,040\).

i) The number of ways is \(3,764,376\).

j) The number of ways is \(109,440\).

Number of tickets \( = 100\) and number of prizes \( = 4\).

**The word "permutation" refers to the act or process of changing the linear order of an ordered set.**

**Formula:**

\(_n{P_r} = \frac{{n!}}{{(n - r)!}}\)

a)

Find the number of ways to select \(4\) people if there is no restriction.

\(\begin{array}{l}P(100,4) = \frac{{100!}}{{(100 - 4)!}}\\P(100,4) = \frac{{100!}}{{96!}}\\P(100,4) = 100 \times 99 \times 98 \times 97\\P(100,4) = 94,109,400\end{array}\)

Hence, the number of ways is \(94,109,400\).

b)

Here, the ticket number \(47\) wins the grand prize and there are three prizes to be awarded.

So, \(n = 99\;,\;r = 3\).

Find the number of ways to select \(3\) people.

\(\begin{array}{l}P(99,3) = \frac{{99!}}{{(99 - 3)!}}\\P(99,3) = \frac{{99!}}{{96!}}\\P(99,3) = 99 \times 98 \times 97\\P(99,3) = 941,094\end{array}\)

Hence, the number of ways is \(941,094\).

c)

Here, the ticket number \(47\) wins the one of the prizes to be awarded.

So, the number of ways to ticket 47 wins the prizes \( = \left( {\begin{array}{*{20}{l}}4\\1\end{array}} \right)\)

For the remaining 3 prizes, \(n = 99\;,\;r = 3\).

Find the number of ways to select \(3\) people.

\(\begin{array}{l}P(99,3) = \frac{{99!}}{{(99 - 3)!}}\\P(99,3) = \frac{{99!}}{{96!}}\\P(99,3) = 99 \times 98 \times 97\\P(99,3) = 941,094\end{array}\)

Find the total number of ways.

Total number of ways \( = 4 \times 941094\)

Total number of ways \( = 3,764,376\)

Hence, the number of ways is \(3,764,376\).

d)

Here, the ticket number \(47\) does not win the prize. So, there are four prizes to be awarded.

Here, \(n = 99\;,\;r = 4\).

Find the number of ways to select \(4\) people.

\(\begin{array}{l}P(99,4) = \frac{{99!}}{{(99 - 4)!}}\\P(99,4) = \frac{{99!}}{{95!}}\\P(99,4) = 99 \times 98 \times 97 \times 95\\P(99,4) = 90,345,024\end{array}\)

Hence, the number of ways is \(90,345,024\).

e)

The ticket number \(47\) wins the one of the prizes.

So, the number of ways to ticket \(47\) wins the prizes \( = \left( {\begin{array}{*{20}{l}}4\\1\end{array}} \right)\)

The ticket number \(19\) wins the one of the prizes.

So, the number of ways to ticket \(19\) wins the prizes \( = \left( {\begin{array}{*{20}{l}}3\\1\end{array}} \right)\)

For the remaining \(2\) prizes, \(n = 98\;,\;r = 2\).

Find the number of ways to select \(2\) people.

\(\begin{array}{l}P(98,2) = \frac{{98!}}{{(98 - 2)!}}\\P(98,2) = \frac{{98!}}{{96!}}\\P(98,2) = 98 \times 97\\P(98,2) = 9506\end{array}\)

Find the total number of ways.

Total number of ways \( = 4 \times 3 \times 9506\)

Total number of ways \( = 114,072\)

Hence, the number of ways is \(114,072\).

f)

The ticket number \(47\) wins the one of the prizes.

So, the number of ways to ticket \(47\) wins the prizes \( = \left( {\begin{array}{*{20}{l}}4\\1\end{array}} \right)\)

The ticket number \(19\) wins the one of the prizes.

So, the number of ways to ticket \(19\) wins the prizes \( = \left( {\begin{array}{*{20}{l}}3\\1\end{array}} \right)\)

The ticket number \(73\) wins the one of the prizes.

So, the number of ways to ticket \(73\) wins the prizes \( = \left( {\begin{array}{*{20}{l}}2\\1\end{array}} \right)\)

For the remaining \(1\) prize, \(n = 97\;,\;r = 1\).

Find the number of ways to select \(1\) people.

\(\begin{array}{l}P(97,1) = \frac{{97!}}{{(97 - 1)!}}\\P(97,1) = \frac{{97!}}{{96!}}\\P(97,1) = 97\end{array}\)

Find the total number of ways.

Total number of ways \( = 4 \times 3 \times 2 \times 97\)

Total number of ways \( = 2328\)

Hence, the number of ways is \(2328\).

g)

The ticket number \(47\) wins the one of the prizes.

So, the number of ways to ticket \(47\) wins the prizes \( = \left( {\begin{array}{*{20}{l}}4\\1\end{array}} \right)\)

The ticket number \(19\) wins the one of the prizes.

So, the number of ways to ticket \(19\) wins the prizes \( = \left( {\begin{array}{*{20}{l}}3\\1\end{array}} \right)\)

The ticket number \(73\) wins the one of the prizes.

So, the number of ways to ticket \(73\) wins the prizes \( = \left( {\begin{array}{*{20}{l}}2\\1\end{array}} \right)\)

The ticket number \(97\) wins the one of the prizes.

So, the number of ways to ticket \(97\) wins the prizes \( = \left( {\begin{array}{*{20}{l}}1\\1\end{array}} \right)\)

Find the total number of ways.

Total number of ways \( = 4 \times 3 \times 2 \times 1\)

Total number of ways \( = 24\)

Hence, the number of ways is \(24\).

h)

None of the people holding ticket \(19\;,\;47\;,\;73\)and \(97\) wins a prize.

Since, \(4\) people do not win the prize.

For the prize, \(n = 96\;,\;r = 4\).

Find the number of ways to select \(4\) people.

\(\begin{array}{l}P(96,4) = \frac{{96!}}{{(96 - 4)!}}\\P(96,4) = \frac{{96!}}{{92!}}\\P(96,4) = 96 \times 95 \times 94 \times 93\\P(96,4) = 79,727,040\end{array}\)

Hence, the number of ways is \(\;79,727,040\).

i)

The number of ways to award a grand prize for \(19\;,\;47\;,\;73\)and \(97\) is \( = \left( {\begin{array}{*{20}{l}}4\\1\end{array}} \right)\)

For the remaining prizes, \(n = 99\;,\;r = 3\).

Find the number of ways to select \(1\) people.

\(\begin{array}{l}P(99,3) = \frac{{99!}}{{(99 - 3)!}}\\P(99,3) = \frac{{99!}}{{96!}}\\P(99,3) = 99 \times 98 \times 97\\P(99,3) = 941,094\end{array}\)

Find the total number of ways.

Total number of ways \( = 4 \times 941094\)

Total number of ways \( = 3,764,376\)

Hence, the number of ways is \(3,764,376\).

j)

The people with ticket \(19\) and \(47\) win the prize.

So, \(n = 4\;,\;r = 2\)

Find the number of ways.

\(\begin{array}{l}P(4,2) = \frac{{4!}}{{(4 - 2)!}}\\P(4,2) = \frac{{4!}}{{2!}}\\P(4,2) = 4 \times 3\\P(4,2) = 12\end{array}\)

The people with ticket \(73\) and \(97\) do not win the prize.

Now, we have to select \(2\) people from the remaining \(96\) people.

So, \(n = 96\;,\;r = 2\).

Find the number of ways.

\(\begin{array}{l}P(96,2) = \frac{{96!}}{{(96 - 2)!}}\\P(96,2) = \frac{{96!}}{{94!}}\\P(96,2) = 96 \times 95\\P(96,2) = 9120\end{array}\)

Find the total number of ways.

Total number of ways \( = 12 \times 9120\)

Total number of ways \( = 109,440\)

Hence, the number of ways is \(109,440\).

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