Question

How many changes can be rung with a peal of 7 bells, the tenor always being last?

Answer 1

Hint: All bells in a peal are different and the position of tenor is fixed. Taking these factors into account permutations is carried out.

As we know that,
Number of ways to arrange n things is $\text{n!}$.
So, the number of changes that can be rung with a peal of $n$ bells is $\text{n!}$.
But here we are given a condition that tenor is always at last.
So, the position of tenor is fixed now.
So, bells left that can still be rearranged are 6 i.e., we can make changes with 6 bells only.
So, the number of changes that can be made with 6 bells is $\text{6!}$.
As we know that $\text{n!}$ is calculated as,
$$\Rightarrow \text{n!}=n\ast (n-1)\ast (n-2).\ast ..........\ast 2\ast 1$$

So, $$\text{6!}=6\ast 5\ast 4\ast 3\ast 2\ast 1=720$$

$$\Rightarrow$$Hence, 720 changes can be rung with a peal of 7 bells, the tenor being last.

Note: Whenever we come up with these types of problems then, we have to only find changes for the objects that are not fixed and that will be $$\text{n!}$$, if n is the number of such objects because if an object is fixed then its position cannot be changed/rearranged.