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The number N=""^(20)C(7)-""^(20)C(8)+""^...

The number `N=""^(20)C_(7)-""^(20)C_(8)+""^(20)C_(9)-""^(20)C_(10)+….. -""^(20)C_(20)` is divisible by :

A

3

B

4

C

7

D

19

Text Solution

Verified by Experts

The correct Answer is:
A, B, C, D
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Similar Questions

Explore conceptually related problems

The sum of the series ""^(20)C_(0)-""^(20)C_(1) +""^(20)C_(2) -""^(20)C_(3)+…+""^(20)C_(10) is :

1.^(20)C_(1)-2.^(20)C_(2)+3.^(20)C_(3)-...-20.^(20)C_(20)

Knowledge Check

  • The number N=""^(20)C_(7)-""^(20)C_(8)+""^(20)C_(9)-""^(20)C_(10)+….. -""^(20)C_(20) is not divisible by :

    A
    3
    B
    7
    C
    11
    D
    19

  • The sum of the series ""^(20)C_(0)-""^(20)C_(1)+""^(20)C_(2)-""^(20)C_(3) + …+ ""^(20)C_(10) is

    A
    `-""^(20)C_(10)`
    B
    `(1)/(2) ""^(20)C_(10)`
    C
    0
    D
    `""^(20)C_(10)`

  • The sum ""^(20)C_(0)+""^(20)C_(1)+""^(20)C_(2)+……+""^(20)C_(10) is equal to :

    A
    `2^(20)+(20!)/(10!)^(2)`
    B
    `2^(19)-1/2*(20!)/(10!)^(2)`
    C
    `2^(19)+""^(20)C_(10)`
    D
    None of these

    • Similar Questions

    Explore conceptually related problems

    The sum of the series 20C_(0)-^(20)C_(1)+^(20)C_(2)-^(20)C_(3)+...-...+^(20)C_(10) is:

    Evaluate: .^(20)C_(5)+^(20)C_(4)+^(21)C_(4)+^(22)C_(4)

    The sum of the series .^(20)C_(0)-.^(20)C_(1)+ .^(20)C_(2)-.^(20)C_(3)+...-.+ .^(20)C_(10) is -

    If ""^(20)C_(r)=""^(20)C_(r-10) , then ""^(18)C_(r) is equal to

    The sum S = ""^(20)C_(2) + 2*""^(20)C_(3) + 3 *""^(20)C_(4) + ...+ 19 * ""^(20)C_(20) is equal to

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