Mathematics, 21.07.2019 23:00 Neko1kat

Achurch has 10 bells in its bell tower. before each church service 3 bells are rung in sequence. no bell is rung more than once. how many sequences are there?

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Answer from: arivalen

To solve this problem you must apply the proccedure shown below:
1. You must apply the following Permutation formula:

nPr=n!/(n-r)!

Where n=8 and r=3

2. Therefore, you have:

n!=8!=8x7x6x5x4x3x2x1=40320
(n-r)!=(8-3)!=5!=5x4x3x2x1=120

3. Then, when you substitute this values into the formula shown above, you obtain:

nPr=n!/(n-r)!
nPr=40320/120=336

4. Therefore, as you can see, the answer is: 336

Answer from: 189652

720

Step-by-step explanation:

Given : A church has 10 bells in its bell tower.

Before each church service 3 bells are rung in sequence.

To Find: How many sequences are there?

Solution:

Since we are given that 3 bells are rung in sequence. no bell is rung more than once.

So, we will use permutation.

Formula : $^nP_r=\frac{n!}{(n-r)!}$

Since A church has 10 bells in its bell tower.

So, n =10

We are also given that Before each church service 3 bells are rung in sequence . So, r =3

So, the number of possible sequence = $^{10}P_3$

= $\frac{10!}{(10-3)!}$

= $\frac{10!}{(7)!}$

= $\frac{10 \times 9 \times 8 \times 7!}{(7)!}$

= $10 \times 9 \times 8$

= 720

Hence there are 720 sequences.

Answer from: Quest

b 1: 45 am

step-by-step explanation:

its that because if it was in the pm it would be 1345

Answer from: Quest

x =±sqrt(1/3)

step-by-step explanation:

x+4/6x=1/x

multiply each side by 6x to get rid of the fractions

6x*(x+4/6x)=1/x*6x

distribute

6x^2 +4 = 6

subtract 4 from each side

6x^2 +4-4 = 6-4

6x^2 =2

divide by 6

6x^2/6 = 2/6

x^2 = 1/3

take the square root of each side

sqrt(x^2) = ±sqrt(1/3)

x =±sqrt(1/3)

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Achurch has 10 bells in its bell tower. before each church service 3 bells are rung in sequence. no...

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