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2 years ago

5. How many ways are there to choose seven cards from a deck of 52?

Mathematics

1 answer:

Fiesta28 [93]2 years ago

7 0

Answer:

674,274,182,400

Step-by-step explanation:

In this particular situation, the first card can be chosen in 52 ways. The second card can be chose in 51 (because when you chose the first card, there were 51 left), you keep doing that seven times

So, 52*51*50*49*48*47*46

Hope this helps.

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Find the probability of rolling a prime number on the first die and even number on the second

Zinaida [17]

Answer:

0.25

Step-by-step explanation:

Prime outcomes on a die: 2,3,5

P(prime) = 3/6 = 1/2

Even outcom: 2,4,6

P(even) = 3/6 = 1/2

P(prime & even)

= P(prime) × P(even)

= ½ × ½

= ¼ or 0.25

4 0

3 years ago

Simplify this expression: 4(x+3)+3(5-x)

prisoha [69]

Answer:

x+27

Step-by-step explanation:

Let's simplify step-by-step.

4(x+3)+3(5−x)

Distribute:

=(4)(x)+(4)(3)+(3)(5)+(3)(−x)

=4x+12+15+−3x

Combine Like Terms:

=4x+12+15+−3x

=(4x+−3x)+(12+15)

=x+27

Hope you find this Useful!

8 0

2 years ago

Read 2 more answers

Subtract -5x^2+2x-5−5x 2 +2x−5 from 7x^2-47x 2 −4.

marta [7]

Answer:

-12x^2+46x-6

Step-by-step explanation:

combine like tens and subtract then from eachother

5 0

2 years ago

A square has a a side lenght of 5 feet what is its area?

goldenfox [79]

Answer:

25 sq feet

Step-by-step explanation:

Length = 5 feet

Area = (Length)²

= 5²

= 25 sq feet

7 0

2 years ago

2. In how many ways can 3 different novels, 2 different mathematics books and 5 different chemistry books be arranged on a books

insens350 [35]

The number of ways of the books can be arranged are illustrations of permutations.

When the books are arranged in any order, the number of arrangements is 3628800
When the mathematics book must not be together, the number of arrangements is 2903040
When the novels must be together, and the chemistry books must be together, the number of arrangements is 17280
When the mathematics books must be together, and the novels must not be together, the number of arrangements is 302400

The given parameters are:

$\mathbf{Novels = 3}$

$\mathbf{Mathematics = 2}$

$\mathbf{Chemistry = 5}$

(a) The books in any order

First, we calculate the total number of books

$\mathbf{n = Novels + Mathematics + Chemistry}$

$\mathbf{n = 3 + 2 + 5}$

$\mathbf{n = 10}$

The number of arrangement is n!:

So, we have:

$\mathbf{n! = 10!}$

$\mathbf{n! = 3628800}$

(b) The mathematics book, not together

There are 2 mathematics books.

If the mathematics books, must be together

The number of arrangements is:

$\mathbf{Maths\ together = 2 \times 9!}$

Using the complement rule, we have:

$\mathbf{Maths\ not\ together = Total - Maths\ together}$

This gives

$\mathbf{Maths\ not\ together = 3628800 - 2 \times 9!}$

$\mathbf{Maths\ not\ together = 2903040}$

(c) The novels must be together and the chemistry books, together

We have:

$\mathbf{Novels = 3}$

$\mathbf{Chemistry = 5}$

First, arrange the novels in:

$\mathbf{Novels = 3!\ ways}$

Next, arrange the chemistry books in:

$\mathbf{Chemistry = 5!\ ways}$

Now, the 5 chemistry books will be taken as 1; the novels will also be taken as 1.

Literally, the number of books now is:

$\mathbf{n =Mathematics + 1 + 1}$

$\mathbf{n =2 + 1 + 1}$

$\mathbf{n =4}$

So, the number of arrangements is:

$\mathbf{Arrangements = n! \times 3! \times 5!}$

$\mathbf{Arrangements = 4! \times 3! \times 5!}$

$\mathbf{Arrangements = 17280}$

(d) The mathematics must be together and the chemistry books, not together

We have:

$\mathbf{Mathematics = 2}$

$\mathbf{Novels = 3}$

$\mathbf{Chemistry = 5}$

First, arrange the mathematics in:

$\mathbf{Mathematics = 2!}$

Literally, the number of chemistry and mathematics now is:

$\mathbf{n =Chemistry + 1}$

$\mathbf{n =5 + 1}$

$\mathbf{n =6}$

So, the number of arrangements of these books is:

$\mathbf{Arrangements = n! \times 2!}$

$\mathbf{Arrangements = 6! \times 2!}$

Now, there are 7 spaces between the chemistry and mathematics books.

For the 3 novels not to be together, the number of arrangement is:

$\mathbf{Arrangements = ^7P_3}$

So, the total arrangement is:

$\mathbf{Total = 6! \times 2!\times ^7P_3}$

$\mathbf{Total = 6! \times 2!\times 210}$

$\mathbf{Total = 302400}$