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Crazy boy [7]
3 years ago
7

In how many ways can 3 red balls , 4 blue balls and

Mathematics
1 answer:
sukhopar [10]3 years ago
8 0

Answer:

1050 ways

Step-by-step explanation:

Total Number of red balls = 5

Total Number of blue balls = 5

Total Number of white balls = 7

Number of red balls selected = 3

Number of blue balls selected = 4

Number of red balls selected = 5

Using the multiplication and combination relation :

nCr * nCr * nCr

Red balls = 5C3

Blue balls = 5C4

White balls = 7C5

Using calculator:

5C3 * 5C4 * 7C5 = 10 * 5 * 21

(10 * 5 * 21) = 1050 ways

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How many eight-digit serial numbers contain three of one digit, two of another digit, two of yet another digit, and one of still
Harlamova29_29 [7]

It is easier to understand the problem if you create a number based on the criteria and then perform the computations. I am going to choose: 111 22 33 4

There are 10 options for the first "1" and only 1 option for the other two 1's

There are 9 remaining options for the first "2" and only 1 option for the other 2

There are 8 remaining options for the first "3" and only 1 option for the other 3

There are 7 remaining options for the "4"

10 x 1 x 1    x      9 x 1     x      8 x 1      x      7

10 x 9 x 8 x 7 = 5,040

Answer: 5,040

6 0
3 years ago
Will give point to first correct answer!!
neonofarm [45]

Answer:

f(x) = 3(0.2)^x

Step-by-step explanation:

The leading coefficient is 3 as  x = 0 gives f(x) = 3.

When x = 1, f(x) = 0.6. so try :

0.6 =3(1.2)^1 = 3.6 so it's not the fiirst choice.

0.6 = 3(0.2)^1 = 0.6 so its last choice.

Check when x = -1:

3(0.2)^-1

= 3/ 0.2

= 15

3 0
2 years ago
Which of the following equation is true ?
VLD [36.1K]
The last one

Hope this helps
3 0
3 years ago
Read 2 more answers
Easy math, but not sure what I did wrong.
Ierofanga [76]

Answer:

8

Step-by-step explanation:

In his 5th year, he took 3 times as many exams as the first year.  So the number of exams taken in the 5th year must be a multiple of 3.

If a₁ = 1, then a₅ = 3.  However, this isn't possible because we need 4 integers between them, and a sum of 31.

If a₁ = 2, then a₅ = 6.  Same problem as before.

If a₁ = 3, then a₅ = 9.  This is a possible solution.

If a₁ = 4, then a₅ = 12.  If we assume a₂ = 5, a₃ = 6, and a₄ = 7, then the sum is 34, so this is not a possible solution.

Therefore, Alex took 3 exams in his first year and 9 exams in his fifth year.  So he took 19 exams total in his second, third, and fourth years.

3 < a₂ < a₃ < a₄ < 9

If a₂ = 4, then a₃ = 7 and a₄ = 8.

If a₂ = 5, then a₃ = 6 and a₄ = 8.

If a₂ = 6, then there's no solution.

So Alex must have taken 8 exams in his fourth year.

4 0
3 years ago
Convert 4/3 to mixed number
elixir [45]
Answer: 1 1/3

All you do is see how many times 3 goes into 4 which is 1 so 1/3 and you add 1 so if added it equals 4 1 1/3

Can I get brainly its okay if no.
6 0
3 years ago
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