Incomplete nCr

ncr.md

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+# Calculating ${n \choose r} \% m$ like a pro
+
+Let's begin with some prerequisites.
+
+## Modular arithmetic
+
+$(a \bmod m)$, also represented as $a \% m =$ remainder left when $a$ is divided by $m$.
+Example: $10 \% 7 = 3$
+
+### Properties of $\%$
+#### Distritbutive over addition & substraction $(\pm)$
+   $$(a\pm b) \% m = \Bigl[ (a \% m ) \pm (b \% m) \Bigr] \% m$$
+   $(8+13)\%7 = \Bigl[(8 \% 7) + (13 \% 7)\Bigr]\%7$
+$21\%7 = [1 + 6]\%7$
+   $0 = 7 \% 7$
+   $0 = 0$
+#### Distritbutive over multiplication $(\times)$
+   $$(a \times b) \% m = \Bigl[ (a \% m ) \times (b \% m) \Bigr] \% m$$
+   $(8 \times 13)\%7 = \Bigl[(8 \% 7) \times (13 \% 7)\Bigr]\%7$
+   $104 \% 7 = [1 \times 6]\%7$
+   $6 = 6 \% 7$
+   $6 = 6$
+    
+- **NOT distributive over $\div$**
+    Unlike $+, -, \times$, the modulo operator does not distribute over division.
+    So, $\left (\dfrac ab \right )\%m \;\;\red \neq\;\; \left[\dfrac{a\%m}{b\%m}\right] \% m$.
+    However, the concept of **Modular Inverse** (discussed later) helps us simplify these calculations.
+
+## Calculating $(n! \bmod m)$ in $\mathcal O(n)$ time
+
+Since $\%$ distributes over $\times$, we can say that
+
+$$(n!) \% m = \Bigl[n (n-1)(n-2) \cdots 1\Bigr] \% m \\[.5em]
+=\overbrace{\underbrace{{\underbrace{\overbrace{n \cdot (n-1)}^{\%m} \cdot (n-2)}_{\%m} \cdots} \cdot 2}_{\% m} \cdot 1}^{\%m}$$
+
+So, we can simply calculate factorial, by taking $\%$  after every multiplication.
+```python
+def fact_mod(n, mod):
+    ans = 1
+    for i in [1 .. n]
+        ans = (ans * i) % mod
+    return ans
+```
+
+## Binary Exponentiation: calculating $(n^p \bmod m)$ in $\mathcal O(\log_2 p)$ time
+```python
+def binary_exponentiation(n, p, mod):
+    if n == 0: return 0
+    if p == 0: return 1
+    ans = binary_exponentiation(n, p//2, mod) # note: integer division
+    ans = ans * ans
+    if p is even:
+        return ans
+    else:
+        return ans * n
+```
+
+
+
+## Fermat's Little Theorem
+If $p$ is prime, then
+$$a^p \equiv a \pmod p$$
+
+## Euler's generalization to Fermat's Little theorem
+
+Iff $a$ and $m$ are co-prime, then
+$$\boxed{a^{\phi(m)} \equiv 1 \pmod m}$$
+
+**Let's break it down:**
+- $a$ is coprime with $m$ iff $gcd(a, m) = 1$, that is, $a$ and $m$ do not have any common factor.
+- If $a \bmod m = b \bmod m$, we say that $a \equiv b \pmod m$
+- $\phi(m)$ is the Euler's Totient Function for $m$
+
+## Euler's Totient Function $\phi(m)$
+The Euler's Totient Function $\phi(m)$ counts the number whole numbers smaller than $m$ which are coprime with $m$
+
+For example, if $m = 30$, then what numbers are coprime with $m$?
+```
+ 1  2  3  4  5  6  7  8  9 10
+11 12 13 14 15 16 17 18 19 20
+21 22 23 24 25 26 27 28 29 30
+-----------------------------
+ 1                 7
+11    13          17    19
+      23                29
+-----------------------------
+count = 8
+```
+
+1. If $m$ is prime,
+    $$\boxed{\phi(m) = m - 1}$$
+2. If $m$ is not prime, then let us factorize $m$
+    Let $m = p_1^{\alpha_1} \cdot p_2^{\alpha_2} \cdot p_3^{\alpha_3} \ldots$
+    Then,
+    $$\boxed{\phi(m) = m \left (1 - \dfrac1 p_1 \right )\left (1 - \dfrac1 p_2 \right )\left (1 - \dfrac1 p_3 \right ) \cdots}$$
+
+Example,
+- if $m = 17$, $\phi(17) = 16$
+- if $m = 30 = 2^1 \cdot 3^1 \cdot 5^1$, then
+    $\phi(30) = 30 \left( 1 - \dfrac 1 2 \right )\left( 1 - \dfrac 1 3 \right )\left( 1 - \dfrac 1 5 \right )$
+    $= 30 \cdot \dfrac12 \cdot \dfrac23 \cdot \dfrac45 = 8$