<p>Given an integer array <code>arr</code>, and an integer <code>target</code>, return the number of tuples <code>i, j, k</code> such that <code>i &lt; j &lt; k</code> and <code>arr[i] + arr[j] + arr[k] == target</code>.</p>

<p>As 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> arr = [1,1,2,2,3,3,4,4,5,5], target = 8
<strong>Output:</strong> 20
<strong>Explanation: </strong>
Enumerating by the values (arr[i], arr[j], arr[k]):
(1, 2, 5) occurs 8 times;
(1, 3, 4) occurs 8 times;
(2, 2, 4) occurs 2 times;
(2, 3, 3) occurs 2 times.
</pre>

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

<pre>
<strong>Input:</strong> arr = [1,1,2,2,2,2], target = 5
<strong>Output:</strong> 12
<strong>Explanation: </strong>
arr[i] = 1, arr[j] = arr[k] = 2 occurs 12 times:
We choose one 1 from [1,1] in 2 ways,
and two 2s from [2,2,2,2] in 6 ways.
</pre>

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

<pre>
<strong>Input:</strong> arr = [2,1,3], target = 6
<strong>Output:</strong> 1
<strong>Explanation:</strong> (1, 2, 3) occured one time in the array so we return 1.
</pre>

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

<ul>
	<li><code>3 &lt;= arr.length &lt;= 3000</code></li>
	<li><code>0 &lt;= arr[i] &lt;= 100</code></li>
	<li><code>0 &lt;= target &lt;= 300</code></li>
</ul>

class Solution:
    def threeSumMulti(self, arr: List[int], target: int) -> int:
        MAXR = 103
        count = [0] * (MAXR)
        MOD = 10 ** 9 + 7
        
        for x in arr:
            count[x] += 1

        '''
        3 number are the same 
        2 number are same, 1 different 
        3 different numbers 
        '''
        total = 0

        # case 1
        i = target // 3
        if i + i + i == target:
            # c * (c + 1) * (c + 2) // 6
            total += math.comb(count[i], 3)
            total %= MOD 

        # case 2
        for i in range(MAXR):
            j = target - i - i

            if j != i and j >= 0 and j < MAXR:
                total += math.comb(count[i], 2) * count[j]
        
        # case 3
        for i in range(MAXR):
            for j in range(i + 1, MAXR):
                k = target - i - j

                if k > j and k >= 0 and k < MAXR:
                    total += count[i] * count[j] * count[k]
                    total %= MOD 

        return total % MOD 

16800