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Question

The number of ways in which $$200$$ different things can be divided into groups of $$100$$ pairs is:


A
(1. 3. 5 ..... 199)
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B
(1012)(1022)(1032)....(2002)
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C
200!2100(100)!
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D
All of these
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Solution

The correct option is C $$\dfrac { 200! }{ { 2 }^{ 100 }\left( 100 \right) ! } $$

We need to form $$100$$ groups having $$2$$ objects each.

Number of possible ways $$=200!$$

Now, we eliminate intrapermutation $$200!$$ by dividing $$2\times 2\times 2......$$ upto $$100=2^{100}$$
200by 

=210so as to account for the intrapermutations of each group.

Since, the group size is the same,the $$200!$$200! we initially got accounts for those cases also in which we are permuting the $$100$$100 groups themselves in $$100!$$ ways.

$$=\dfrac{200!}{2^{100}\times 100!}$$


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