jeqcho's blog

By jeqcho, history, 9 months ago, In English

Sometimes, you are asked to calculate the combination or permutation modulo a number, for example $$$^nC_k \mod p$$$. Here I want to write about a complete method to solve such problems with a good time complexity because it took me a lot of googling and asking to finally have the complete approach. I hope this blog can help other users and save their time when they solve combinatorics problem in Codeforces.

Example Problem

Find the value of $$$^nC_k$$$, ($$$1 \leq n,k \leq 10^6$$$). As this number can be rather large, print the answer modulo $$$p$$$. ($$$p = 1000000007 = 10^9 + 7$$$)

Combination (binomial coefficients)

$$$^nC_k$$$ means how many ways you can choose $$$k$$$ items from an array of $$$n$$$ items, also denoted as $$$\binom{n}{k}$$$. This is also known as binomial coefficients. The formula for combination is

$$$ ^nC_k = \frac{n!}{k!(n-k)!} $$$

Sometimes, the denominator $$$k!(n-k)!$$$ is very large, but we can't modulo it since modulo operations can't be done independently on the denominator. $$$\frac{n!}{k!(n-k)!} \mod p \neq \frac{n! \mod p}{k!(n-k)! \mod p}$$$. Now I will introduce the modular multiplicative inverse to solve this problem.

Modular multiplicative inverse

The modular multiplicative inverse $$$x$$$ of $$$a$$$ modulo $$$p$$$ is defined as

$$$ a \cdot x \equiv 1 \pmod p $$$

Here, I will replace $$$x$$$ with $$$\text{inv}(a)$$$, so we have

$$$ a \cdot \text{inv}(a) \equiv 1 \pmod p $$$

Getting back to the formula for combination, we can rearrange so that

$$$ ^nC_k = n! \cdot \frac{1}{k!} \cdot \frac{1}{(n-k)!} $$$

Here, we can use $$$\text{inv}(a)$$$ as follows

$$$ ^nC_k \equiv n! \cdot \text{inv}(k!) \cdot \text{inv}((n-k)!) \pmod p $$$

Now we can distribute the modulo to each of the terms by the distributive properties of modulo

$$$ ^nC_k \mod p = n! \mod p \cdot \text{inv}(k!) \mod p \cdot \text{inv}((n-k)!) \mod p $$$

Now I will discuss on how to calculate $$$\text{inv}(a)$$$

Fermat's Little Theorem

You can easily remember this theorem. Let $$$a$$$ be an integer and $$$p$$$ be a prime number,

$$$ a^p \equiv a \pmod p $$$

It is helpful to know that the $$$p$$$ in the problem ($$$10^9 + 7$$$) is indeed a prime number! We can rearrange the equation to get

$$$ a^{p-1} \equiv 1 \pmod p $$$

Looking back at our equation for $$$\text{inv}(a)$$$, both equations equate to 1, so we can equate them as

$$$ a \cdot \text{inv}(a) \equiv a^{p-1} \pmod p $$$

We can rearrange the equation to get

$$$ \text{inv}(a) \equiv a^{p-2} \pmod p $$$

We now have a direct formula for $$$\text{inv}(a)$$$. However, we cannot use the pow() function to calculate $$$a^{p-2}$$$ because $$$a$$$ and $$$p$$$ is a large number (Remember $$$1 \leq n,k \leq 10^6$$$) ($$$p = 10^9 + 7$$$). Fortunately, we can solve this using modular exponentiation.

Modular Exponentiation

To prevent integer overflow, we can carry out modulo operations during the evaluation of our new power function. But instead of using a while loop to calculate $$$a^{p-2}$$$ in $$$O(p)$$$, we can use a special trick called exponentiation by squaring. Note that if $$$b$$$ is an even number

$$$ a^b = (a^2)^{b/2} $$$

Every time we calculate $$$a^2$$$, we reduce the exponent by a factor of 2. We can do this repeatedly until the exponent becomes zero where we stop the loop. This will give us a time complexity of $$$O(\log p)$$$ to calculate $$$a^{p-2}$$$ because we halve the exponent in each step. For the case when $$$b$$$ is odd, we can use the property

$$$ a^b = a^{b-1} \cdot a $$$

We then store the trailing $$$a$$$ into a variable. Then $$$b-1$$$ is even and we can proceed as previously stated. We can repeatedly apply these two equations to calculate $$$a^{p-2}$$$. Here I will show you the implementation of this modified powmod() function to include modulo operations. ll is defined as long long

ll powmod(ll a, ll b, ll p){
    a %= p;
    if (a == 0) return 0;
    ll product = 1;
    while(b > 0){
        if (b&1){    // you can also use b % 2 == 1
            product *= a;
            product %= p;
        a *= a;
        a %= p;
        b /= 2;    // you can also use b >> 1
    return product;

Then we can finally implement the $$$\text{inv}(a)$$$ function simply as

ll inv(ll a, ll p){
    return powmod(a, p-2, p);

Then, finally, we can implement $$$^nC_k$$$ as

ll nCk(ll n, ll k, ll p){
    return ((fact[n] * inv(fact[k], p) % p) * inv(fact[n-k], p)) % p;

We used the dp-approach for factorial where the factorial from 1 to n is pre-computed and stored in an array fact[].

Time complexity

  • Pre-computation of factorial: $$$O(n)$$$
  • Calculation of $$$^nC_k$$$, which is dominated by modular exponentiation powmod: $$$O(\log p)$$$
  • Total: $$$O(n + \log p)$$$


Problems for you

Please comment below if you know similar problems.

I hope this blog will help you in your competitive programming journey.

Stay safe and thank you for reading.

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9 months ago, # |

Auto comment: topic has been updated by jeqcho (previous revision, new revision, compare).

9 months ago, # |

Good tutorial! It's worth mentioning that you can find modular inverse using Extended Euclidean Algorithm in $$$O(\log{p})$$$, too.

Also, if $$$n$$$ and $$$k$$$ are small, you can calculate binomial coefficients with DP in $$$O(nk)$$$ without modular inverse.

And here are some problems:

8 months ago, # |

// you can also use (b >>= 1).

for correction.

5 weeks ago, # |

Thank you. This tutorial was very helpful and easy to understand.

5 weeks ago, # |

Thank you for tutorial!