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2 years ago

12

How many ways can five people, a,b,c,d, and e, sit in a row at a movie theater if c must sit to the right of but not necessarily

next to b?

2 answers:

Snezhnost [94]2 years ago

8 0

Answer:

a,b,c,d,e

a,b,d,c,e

a,b,d,e,c

b,a,d,e,c

b,d,a,e,c

b,d,e,a,c

hichkok12 [17]2 years ago

7 0

A,b,c,d,e
a,b,d,c,e
a,b,d,e,c
b,a,d,e,c
b,d,a,e,c
b,d,e,a,c

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Mixed numbers from 3 to 7 with an interval of 1/3

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4 1/3 there you go hope I helped

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3 years ago

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Solve the following inequality: –1 + 6(–1 – 3x) > –39 – 2x.

insens350 [35]

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2 years ago

Find the 4th term (x-y)^12

maria [59]

Answer:

The fourth term of the expansion is -220 * x^9 * y^3

Step-by-step explanation:

Question:

Find the fourth term in (x-y)^12

Solution:

Notation: "n choose k", or combination of k objects from n objects,

C(n,k) = n! / ( k! (n-k)! )

For example, C(12,4) = 12! / (4! 8!) = 495

Using the binomial expansion formula

(a+b)^n

= C(n,0)a^n + C(n,1)a^(n-1)b + C(n,2)a^(n-2)b^2 + C(n,3)a^(n-3)b^3 + C(n,4)a^(n-4)b^4 +....+C(n,n)b^n

For (x-y)^12, n=12, k=3, a=x, b=-y, and the fourth term is

C(n,3)a^(n-3)b^3

=C(12,3) * x^(12-3) * (-y)^(3)

= 220*x^9*(-y)^3

= -220 * x^9 * y^3

3 0

3 years ago

To play the lottery in a certain state, a person has to correctly select 5 out of 45 numbers, paying a $1 for each five-number s

defon

There are 45 numbers out of which 5 numbers are required to win an enormous sum of money

No of ways in which 5 numbers can be selected out of 45 numbers.

$C_{45}^5=\frac{45!}{40!5!} =1221759$

out of which there is only combination foe contest winning

$P=\frac{1}{1221759}$

When 100 different tickets are bought then

Probability of winning: $\frac{100}{1221759}$

Probably of winning≈0,000082

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P.S. Hello from Russia :^)

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