I was reading a book where the author calculated the final price using a formula I can't understand how it was obtained. I would like to know how it was derived and how it could be generalized.

Example: Price is $800$ USD. Discounts applied are in the following sequence: $20/100$, $5/100$ and $5/100$ again.

The final price $$= \frac{800 * 4 * 19 * 19}{5 * 20 * 20}=577.60$$.

My question is how was the above formula derived? How could it be generalized, say given an item price to be $p_0$, a set of discounts ($d_1$,$d_2$,...,$d_{n-1}$) are applied in sequence to this price to get the final price $p_n$.

Note: I know how to calculate the same answer in different ways, but I found this formula not so obvious to me.

  • 2
    $\begingroup$ $20/100 = 1/5, \; 5/100 = 1/20$, can you relate theses figures to those used in the final formula ?, $\endgroup$ Commented May 17 at 5:37
  • $\begingroup$ "20% of'' means ''$\frac{20}{100}$ multipled by''. $\endgroup$
    – Paul
    Commented May 17 at 6:24
  • $\begingroup$ @True, I can produce a formula indeed but (sadly) it does not look like the one above...at all! $\endgroup$
    – NoChance
    Commented May 17 at 7:57
  • $\begingroup$ @Paul, thanks, I get this. $\endgroup$
    – NoChance
    Commented May 17 at 7:57

1 Answer 1


When we talk about a "discount of $d$," where $0 < d < 1$, what we mean is that if the original price is $P$, then we are reducing the price by $P \times d$. So for instance, if $P = 100$, and the discount is $d = 0.25 = \frac{1}{4} = 25\%$, that means we are reducing the price by $P \times d = 100 \times 0.25 = 25$. The discounted price is $$P - P \times d = 100 - 25 = 75.$$

We can shorten this calculation into a single step: if $d$ is the discount rate, then $P(1-d)$ is the discounted price when $P$ is the original price.

Consequently, if we have a set of discount rates $d_1, d_2, \ldots, d_{n-1}$, and an original price $p_0$, the price after the application of all of the discounts is $$p_n = p_0 (1 - d_1)(1 - d_2) \cdots (1 - d_{n-1}).$$ This is precisely how the calculation was done in the example you cited: $$800 (1 - 0.20)(1 - 0.05)(1 - 0.05).$$

  • $\begingroup$ Clear now. I was under the assumption that $p_{n-k}$ would be required which I could not simplify. However, your answer is elegant. Thank. Also, the fact that $(1 - 0.05)$ can be expressed as $(\frac{19}{20})$ was not obvious at fist sight. $\endgroup$
    – NoChance
    Commented May 17 at 8:32

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