Prove that 2C(n, 2) + n^2 = C(2n, 2)for n>=2?

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  • atsuo
    Lv 6
    2 weeks ago

    Think that there are n boys and n girls (n≧2) ,

    and you choose 2 persons out of them .

    If you choose 2 boys , then C(n,2) ways exist .

    If you choose 2 girls , then C(n,2) ways exist .

    If you choose 1 boy and 1 girl , then C(n,1)*C(n,1) = n^2 ways exist .

    So total of 2C(n,2) + n^2 ways exist .

    On the other hand , there are C(2n,2) ways to choose 2 persons out of 2n persons .

    Therefore , 2C(n,2) + n^2 = C(2n,2) .

  • 3 weeks ago

    C(n, 2) = n!/(2!(n-2!) = (n²-n)/2

    C(2n, 2) = ((2n)²-2n)/2 = (4n²-2n)/2 = 2×(n²-n)/2 + (2n²)/2 = 2×C(n, 2) + n²

    This is true where nC2 exist which is n ≥ 2

  • Pope
    Lv 7
    3 weeks ago

    For any integer r ≥ 2,

    C(r, 2) = r!/[2!(r - 2)!] = r(r - 1)(r - 2)!/[2!(r - 2)!] = r(r - 1)/2! = r(r - 1)/2

    2C(n, 2) + n²

    = 2[n(n - 1)/2] + n²

    = 2n² - n

    = (4n² - 2n)/2

    = 2n(2n - 1)/2

    = C(2n, 2)

  • 3 weeks ago

    Note, C(n, r) = n!/(n-r)!r!, where n ≥ r,

    2C(n, 2) + n^2 = C(2n, 2),

    LHS = 2[n!/2(n-2)!] + n^2

    LHS = n(n-1) + n^2 = n(2n-1),

    RHS = (2n)!/(2n-2)!2! = 2n(2n-1)/2 = n(2n-1)

    ∴ LHS = RHS and the result is OK.

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