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Me asked in Science & MathematicsMathematics · 6 years ago

If x, y, z and t are odd natural numbers such that x + y + z + t = 20, then find the number of values of ordered quadruplet(x, y, z and t).?

Permutations and combinations problem

I've done 20-10 for odd numbers.

Then 10 + 4 - 1 C 10 ie 13 C 10.

But the ans is coming wrong. What am I doing wrong?

2 Answers

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  • ??????
    Lv 7
    6 years ago
    Favorite Answer

    Transform the problem by writing

    x = 2 x' + 1

    y = 2 y' + 1

    z = 2 z' + 1

    t = 2 t' + 1

    => x + y + z + t = 2 (x' + y' + z' + t') + 4 = 20

    => x' + y' + z' + t' = 8

    This a stars and bars problem.

    We have 8 stars to divide under 4 groups.

    So we put 3 bars for the subdivision.

    => C(8 + 3, 3) = C(11,3) = 11x10x9/(1x2x3) = 11x5x3 = 165 possibilities.

    The answer is 165.

  • Nick
    Lv 6
    6 years ago

    (Note that my y and z are indeterminates and not your x,y,z,t)

    The generating function for odd values of each of variable is z^1 + z^3 + z^5 + z^7 + ... = z/(1-z^2) so the ogf for the system is:

    z^4/(1-z^2)^4

    since there is a known expansion:

    y^k/(1-y)^(k+1) = sum(C(n,k)y^n)

    replace y with z^2 and k with 3 we have:

    z^6/(1-z^2)^4 = sum(C(n,3)(z^2)^n)

    z^2(z^4/(1-z^2)^4) = z^2*(sum(C(n,3)z^(2(n-1))))

    z^4/(1-z^2)^4 = sum(C(n,3)z^(2(n-1)))

    z^4/(1-z^2)^4 = sum(C(m/2 +1,3)z^m)

    the appropriate coefficient in this summation is that of the z^20 term, so m=20:

    C(20/2 + 1,3) = C(11,3) = 11*10*9/(3*2) = 165 <----

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