There were 5 seats on the bus. In how many ways could the seats be occupied if there were four children?

Mathematics

1 answer:

Lana71 [14]3 years ago

8 0

Answer:

5 choose 4

$=\frac{5!}{4!\left(5-4\right)!}$ = 5

Step-by-step explanation:

You might be interested in

Does a number go into 49 evenly

Gnoma [55]

<span>We are asked if a number goes into 49 evenly.

First, let's explore what it means to "go in evenly."

A number (call it A) goes into another number (Call it B) if when we divide B by A there is no remainder. As an example, 5 goes into 10 evenly. This is because 10/5 = 2. There is no remainder. However, 6 does not go into 10 evenly. This is because 10/6 = 1 remainder 4. Since there is a remainder when we divide 10 by 6 we know 6 does not go evenly into 10.

If we go back to our example of 10 and 5. Since 5 goes evenly into 10 we call 5 a divisor (or factor) or 10. We call 10 a multiple of 5. Let's list all of the numbers that go evenly into 10: 1, 2 5, 10. These are the factors of 10.

Notice that 1 and the number itself are always factors of a number. What the question is asking us is whether there are any factors of 49. We know that 1 and 49 go evenly into 49, but in these types of questions we are looking for factors other than 1 or the number because those always work.

Do we have to test all of the numbers between 1 and 49? Do we have to divide 49 by all the number between 1 and 49 to find those that leave no remainder? Well, we could BUT you might have noticed that factors come in pairs so it's enough to check half the numbers, that is the numbers from 1 to 25.

When we divide 49 by one of these numbers and get a zero reminder the answer to our division problem is also a factor. If you do this you will see that 49 / 7 = 7 with no remainder. So 7 is a number that goes evenly into 49 (so is the other 7 but we don't have to write it twice).

You might have some familiarity with multiplication facts that allows you to get to the answer 7 (maybe you remember that 7 * 7 = 49) without testing the numbers one by one but testing them will always work.</span>

3 0

3 years ago

Somebody help! How do I do this?

nlexa [21]

Answer:

${ \tt{3y - 9 = { \blue{ \boxed{3}}(y) - { \blue{ \boxed{3}}(3)}}}} \\ \\ { \tt{3y - 9 = 3({ \blue{ \boxed{{ \tt{y}}}} - { \blue{ \boxed{3}})}}}} \\$