Given two positive numbers ‘n’ and ‘m’ (n <= m) which represent total number of items of first and second type of sets respectively. Find total number of ways to select at-least one pair by picking one item from first type(I) and another item from second type(II). In any arrangement, an item should not be common between any two pairs.
Note: Since answer can be large, output it in modulo 1000000007.
Input: 2 2 Output: 6 Explanation Let's denote the items of I type as a, b and II type as c, d i.e, Type I - a, b Type II - c, d Ways to arrange one pair at a time 1. a --- c 2. a --- d 3. b --- c 4. b --- d Ways to arrange two pairs at a time 5. a --- c, b --- d 6. a --- d, b --- c Input: 2 3 Output: 12 Input: 1 2 Output: 2
The approach is simple, we only need the combination of choosing ‘i‘ items from ‘n‘ type and ‘i‘ items from ‘m‘ type and multiply them(Rule of product) where ‘i‘ varies from 1 to ‘n‘. But we can also permute the resultant product in ‘i’ ways therefore we need to multiply with i!. After that take the sum(Rule of sum) of all resultant product to get the final answer.
Time complexity: O(m*log(mod))
Auxiliary space: O(m)
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