Basically, Division is the inverse of multiplication, see this instance:

5 × 4,
This basically means, adding five four times,
5, 5, 5, 5 added together is 20;

If division is the inverse of multiplication and multiplication comes from repeated addition, how can we interpret division as repeated subtraction?
How can we interpret this as the statements in bold above?

That's the level 1 of confusion, the next is, while taking the word PEPPER:
Total arrangement is given by the formula:

How can we interpret the result as above, i.e. with the division interpreted as a series of repeated subtraction?
EDIT: And how do we relate this with the fact that division is breaking things up equally with respect to the denominator? And in this case:
Why isn't 6!-(3!2!)used?

  • 1
    $\begingroup$ I would not recommend thinking of operations and their inverses as analogous to addition and subtraction, as they often will not work the same. For example, every real number has an additive inverse however there is no multiplicative inverse for the number zero. If you insist however, then for the specific case of integers $n$ and $k$ where $k\mid n$, then $n/k=x$ would be like interpreting $n$ as $\underbrace{x+x+\dots+x}_{k~\text{copies}}$ and keeping only one copy of $x$ while discarding all of the rest. It is not really "repeated subtraction" but rather, removal of identical copies. $\endgroup$ – JMoravitz Jan 4 '17 at 5:13

If we want to do some division using subtraction. We have to use successive subtraction until we got zero. And the number of time you subtract is the answer.

In your first case,

20 - 4 = 16

16 - 4 = 12

12 - 4 = 8

8 - 4 = 4

4 - 4 = 0

So we subtract 4 five times so answer is 5

In second case 6! = 720 and 2!3! = 12

So we have to subtract 12 sixty times to get zero so answer is 60.

Edit -

We can use any operation on factorials directly.

So you have to do 720 - 12 repeatedly.

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If you want to interpret "$20$ divided by $4$" in terms of repeated subtraction, what you would do is repeatedly subtract $4$ from $20$ until you get $0$:

$$16, 12, 8, 4, 0$$

The number of times you had to subtract $4$ is exactly the quotient $\frac{20}{4}$. This same principle carries over to the quotient $\frac{6!}{3!2!}$: you repeatedly subtract $3!2!$ from $6!$ until you get zero, and the number of times you subtracted it is the quotient.

The same principle sort of works when $a, b$ are positive integers, but $\frac{a}{b}$ is not an integer. Let's say you want to interpret the division of $27$ by $4$ as repeated subtraction. What you would do is repeatedly subtract $4$ from $27$, until you eventually get a number between $0$ and $3$. The number of times $q$ that you subtracted $4$ to get to that point, and the number $r$ between $0$ and $3$ that you get, satisfy $q = 6$ and $r = 3$. Then

$$q + \frac{r}{4} = 6 + \frac{3}{4}$$

is equal to the quotient $\frac{27}{4}$.

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  • $\begingroup$ How do we relate this with also the concept of division as making equal parts? Division also means, getting equal number of parts as the denominator, How do we relate these two processes? $\endgroup$ – mathmaniage Jan 4 '17 at 5:23
  • $\begingroup$ please see my edits. $\endgroup$ – mathmaniage Jan 4 '17 at 5:28

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