Number of Music Playlists
<p>Your music player contains <code>n</code> different songs. You want to listen to <code>goal</code> songs (not necessarily different) during your trip. To avoid boredom, you will create a playlist so that:</p>

<ul>
	<li>Every song is played <strong>at least once</strong>.</li>
	<li>A song can only be played again only if <code>k</code> other songs have been played.</li>
</ul>

<p>Given <code>n</code>, <code>goal</code>, and <code>k</code>, return <em>the number of possible playlists that you can create</em>. Since the answer can be very large, return it <strong>modulo</strong> <code>10<sup>9</sup> + 7</code>.</p>
<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> n = 3, goal = 3, k = 1
<strong>Output:</strong> 6
<strong>Explanation:</strong> There are 6 possible playlists: [1, 2, 3], [1, 3, 2], [2, 1, 3], [2, 3, 1], [3, 1, 2], and [3, 2, 1].
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> n = 2, goal = 3, k = 0
<strong>Output:</strong> 6
<strong>Explanation:</strong> There are 6 possible playlists: [1, 1, 2], [1, 2, 1], [2, 1, 1], [2, 2, 1], [2, 1, 2], and [1, 2, 2].
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> n = 2, goal = 3, k = 1
<strong>Output:</strong> 2
<strong>Explanation:</strong> There are 2 possible playlists: [1, 2, 1] and [2, 1, 2].
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>0 &lt;= k &lt; n &lt;= goal &lt;= 100</code></li>
</ul>
Math; Dynamic Programming; Combinatorics
function numMusicPlaylists(n: number, goal: number, k: number): number {
    const mod = 1e9 + 7;
    const f = new Array(goal + 1).fill(0).map(() => new Array(n + 1).fill(0));
    f[0][0] = 1;
    for (let i = 1; i <= goal; ++i) {
        for (let j = 1; j <= n; ++j) {
            f[i][j] = f[i - 1][j - 1] * (n - j + 1);
            if (j > k) {
                f[i][j] += f[i - 1][j] * (j - k);
            }
            f[i][j] %= mod;
        }
    }
    return f[goal][n];
}
