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Question

# Let N=2200(200C0)(200C100)−2198(200C1)(199C99)+2196(200C2)(198C98)+⋯+(200C100)(100C0) then -

A
N is equal to the coefficient of x100 in (1x)200
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B
The number of zeros at the end of N is only one.
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C
N=3101×k, where k is an integer
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D
N is equal to the number of ways of distributing 200 different objects among 2 persons equally.
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Solution

## The correct option is C N=3101×k, where k is an integer N=100∑r=0(−1)r.4100−r⋅200Cr⋅200−rC100−r =100∑r=0(−1)r.4100−r×200!r!(200−r)!×(200−r)!(100−r)!100!×100!100! N= 200C100 .(4−1)100 N= 200C100 .3100 no. of zero's in 200C100 is equal to exponent of 5. 200!100!100!=(40+8+1)(2×(20+4+0)) = exp. of 5 → 49−48=1 Let's find the exponent of 3 in 200C100 200C100=200!100!×100!exponent of 3 in200C100=exponent of 3 in 200!exponent of 3 in 100!×100!=97−96=1⇒ 200C100 3100=k×3101 the number of ways of distributing 200 different objects among 2 persons equally =200!100! 100!×12!×2!= 200C100 the coefficient of x100 in (1−x)200= 200C100

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