Inventory: An Analysis on Separation An inventory consists of a list of $115$ items, each marked "available" or "unavailable." There are $60$ available items. Show that there are at least two available items in the list exactly four items apart.

I think the problem can be worded this way:

Every permutation of a $115$-bit string with exactly sixty $1$'s will each be found to have two $1$'s separated by exactly four $1$'s or $0$'s.

 A: Hint: Number the items $1$ to $115$. 
Divide the numbers into $5$ groups. (i) the ones diviible by $5$; the ones that leave a remainder of $1$ on division by $5$: and so on up to the ones that leave a remainder of $4$ on division by $5$. 
Count how many there are in each group. Show that if we have $60$ items, there are "too many" items in at least one of the groups. 
By too many items, think of a much smaller group like $1,6,11,16,21,26,31,36,41,46$. Note that if we take $6$ items from this group, at least two will be $4$ apart. 
Added: Each of the congruence classes modulo $5$ has $21$ members, and there are $5$ of them. So if we have $60$ numbers, one of the congruence classes at least must have $12$ or more representatives in the bunch of $60$.  They are sepaated by $5$ or $10$ or more. We cannot arrange for them to be all separated by $10$ or more, since there are too many.
A: Line up the items. $60$ are available and $55$ are unavailable. Of the first $111$ items in the list there are at least $56$ available. Let us consider the proposition false, namely that there are not at least two available items exactly four items apart, but this would mean that the item for positions down the list of each available item is unavailable, but then there are at least $56$ items that have an item $4$ items down the list, but there are $55$ items  unavailable in the whole list, so there is a contradiction.
