Consider the equation $$z_1 + z_2 + z_3 + z_4 + z_5 + z_6 = k$$ For: $i = 1, \dotsc,6$ $z_i$ is a positive natural number and they must satisfy the following: \begin{align} z_1 & \ge 4 \\ z_2 & \ge 10 \\ z_3 & \ge 12 \\ z_4 & \ge 18 \\ z_5 & \ge 6 \\ z_6 & \ge 30 \end{align}
(a) For what values of $k$ are there no solutions?
(b) Determine the number of solutions for other values of $k$
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mvw
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wout konings
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5Please add what you have tried so far to the question so that people on the site can help you solve your problem. – Apurv Jun 13 '16 at 09:33
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Can you not say anything at all about (a), wout? Can you not see that there's no solution for $k=-17$? for $k=100\sqrt2$? for $k=20$? – Gerry Myerson Jun 13 '16 at 09:53
2 Answers
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(a) The minimal $k$ value is for $z = b$: $$ k = 4 + 10 + 12 + 18 + 6 + 30 = 80 $$ For smaller $k$ values there is no solution.
(b)
For $k = 80 + 1$ we have $6$ components to distribute the additional $1$ to.
For $k = 80 + 2$ we have $6$ possible choices for the first additional $1$ and again $6$ possible choices for the second additional $1$. However some of the choices lead to an indistinguishable result, like $(1,2)$ and $(2,1)$, so we have $(6^2 - 6)/2 + 6 = 15 + 6 = 21$ choices.
For $k = 80 + n$, $n \in \mathbb{N}$, it is the general combinatorial problem of putting $n$ identical objects into $6$ bins. See e.g. here. This gives $\binom{n+6-1}{6-1} = \binom{n+5}{5}$ choices.
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$(a)$ $^{k-(4+10+12+18+6+30)+(n-1)}C_{(n-1)}=^{(k-81+6)}C_{(6-1)}=^{(k-75)}C_5 \neq 0 $
For minimum one solution : $^{(k-75)}C_5=1 \implies (k-75)=5, \implies k=80$.
So, solution is not possible only when $k<80$ for given conditions.
$(b)$ When restriction is not applied the number of solutions $=^{(k+(n-1))}C_{(n-1)}$.
If these conditions applied :
Value of $k$ should be $k\geq 80$ for atleast one solution and $k$ should be integer.
In how many ways can 20 identical balls be distributed into 4 distinct boxes subject?
Mithlesh Upadhyay
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