Question

A church 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?

Answer 1

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 2

Answer:

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.