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

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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,
\[ \Rightarrow {\text{n!}} = n*(n - 1)*(n - 2).*..........*2*1\]

So, \[{\text{6!}} = 6*5*4*3*2*1 = 720\]

\[ \Rightarrow \]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.