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in Binomial Theorem by (70.4k points)
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In , Example 28 , the number of ways of choosing A
and B such that A and B have equal number of elements, is
A. `2^(n)`
B. `3^(n)`
C. `(2^(n))^(2)`
D. `""^(2n)C_(n)`

1 Answer

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by (71.0k points)
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Best answer
Correct Answer - d
If A contains `r( 0 le r le n)` elements, then B must
also contain r elements. In this case the number of ways of
choosing A and B is` ""^(n)C_(r) xx""^(n)C_(r) = (""^(n)C_(r))^(2)`. But, r can very from 0 to
n. So, the number of ways of choosing A and B is
`sum_(r=0)^(n) (""^(n)C_(r))^(2) = sum_(r=0)^(n) C_(r)""^(2)= C_(0)""^(2)+C_(1)""^(2) + ...+ C_(n)""^(2) = ""^(2n)C_(n)`.

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