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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
- ??????Lv 76 years agoFavorite 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.
- NickLv 66 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 <----