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jok3333 [9.3K]
3 years ago
9

A popular band wants to sell $44,856 worth of tickets on its upcoming tour. If each ticket cost $24, how many tickets will the b

and have to sell to meet its goal?
Mathematics
1 answer:
liraira [26]3 years ago
6 0
They will have to sell 1869 tickets.
What you basically do to find an answer is divide 24 in 44,858.
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One is c two is also c three is c and the fourth is d
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\text{C. }60

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<u>Question from image</u>:

"The digits 1, 2, 3, 4, and 5 can be formed to arrange 120 different numbers. How many of these numbers will have the digits 1 and 2 in increasing order? For example, 14352 and 51234 are two such numbers."

Let's start by taking a look with the number 1. There are four possible places 1 could be, because there needs to be space for the 2 after it.

Checkmarks mark where the 1 can be, the <em>x</em> marks where it cannot be.

\underline{\checkmark}\:\underline{\checkmark}\:\underline{\checkmark}\:\underline{\checkmark}\:\underline{X}

Let's start with the first position:

\underline{\checkmark}\:\underline{\#}\:\underline{\#}\:\underline{\#}\:\underline{\#}

There are four places the 2 can be. For each of these four places, we can arrange the remaining 3 digits in 3!=6 ways. Therefore, there are 4\cdot 6=\boxed{24} possible numbers when 1 is the first digit of the number.

Continue this process with the remaining possible positions for 1.

Second position:

\underline{X}\underline{\checkmark}\:\underline{\#}\:\underline{\#}\:\underline{\#}

There are three places the 2 can be, since the 2 must be behind the 1. For each of these three places, the remaining 3 digits can be arranged in 3!=6 ways. Therefore, there are 3\cdot 6=\boxed{18} possible numbers when 1 is the second digit of the number.

The pattern continues. Next there will be 2 places to place the 2. For each of these, there are 3!=6 ways to rearrange the remaining 3 digits for a total of 2\cdot 6=\boxed{12} possible numbers when 1 is the third digit of the number.

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