Discrete Mathematics By Kenneth H Rosen Counting

Question

One Hundred tickets, numbered $1,2,3,...,100$, are sold $100$ 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

1. the people holding tickets $19$ and $47$ both win prizes?
2. the people holding tickets $19,47,$ and $73$ all win prizes?

Comments

is the answer for a) 114072

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And the answer for b) 2328?

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Yes to give prize for 19 Total ways =4 to give prize for 47 Total ways =3  so for rest 2 prizes 98 ,97

1)4*3*98*97=114072

similarly for second

2)4*3*2*97=2328

Answer 1

There are $100$ different people & $100$ different tickets. And among the tickets, there are $4$ tickets by getting those one would get a prize.

Now the criteria is : $\color{violet}{\text{a) The people holding tickets 19 and 47 both win prizes.}}$

The possibility of $19$ and $47$ (ticket number) both get prizes = $4\times3 \hspace{0.1cm}ways$  $\qquad  \big[\text{∵ As there are 4 prizes}\big]$

Remaining Prizes = $(4-2) = 2$

These $2$ prizes can be divided among the $(100-2)=98$ people in $\{98\times 97\hspace{0.1cm} ways\}$

So,$\color{green}{\text{ Total number of ways in which the people holding tickets}}$ $\color{red}{19}$ and $\color{red}{47}$ $\color{green}{\text{both win prizes}}$ $\color{blue}{=4\times 3 \times 98 \times 97 = 114072\hspace{0.1cm} ways}$

Now, the second criteria is$\color{violet}{\text{ b) The people holding ticket number 19, 47, 73 all win prizes}}$?

∴ The ticket number $19, 73, 47$ all win prizes in $\{4\times 3 \times 2\hspace{0.2cm} ways\}$

And the remaining $\{(4-3)=1\}$ tickets can be distributed in remaining $\{(100-3) =97 \}$ people in $97$ ways.

So, $\color{green}{\text{The total number of ways in which the people holding tickets}}$ $\color{gold}{\text{19,47,}}$ $\color{green}{and}$ $\color{gold}{73}$ $\color{green}{\text{all win prizes is }}$ $=\color{maroon}{4 \times 3 \times 2\times 97 = 2328 \hspace{0.1cm}ways}$