High computation in probability 
Six men and some number of women stand in a line in random order. Let $p$ be the probability that a group of at least four men stand together in the line, given that every man stands next to at least one other man. Find the least number of women in the line such that $p$ does not exceed 1 percent.

Obviously:
$$\overbrace{P(\text{at least four in a line})}^{P_T} = P(\text{4 in line}) + P(\text{5 in line}) + P(\text{6 in line})$$
But first the counting Aspect.
We have lined up:
$$MMMMMMWWWWWW...$$
To get exactly $2$ groups of $4$ men,  we have choices:
$$MMWMMMM \space \text{AND} \space MMMMWMM$$ But I see that we could also have:
$$MMWWWWWWWWWMMMM \space \text{AND} \space MMMMWWWWWWWWMM$$
So suppose we have $x$ women, so $x*W$ total. 

Then I am confused.
To get two groups of four men together, how many ways are there? I would say $2$ but that doesnt use $x$ or $W$ then?

 A: Suppose there are $x$ women. I assume them all to be the same, like you seem to do in the question. 
Case 1. If there is a block of exactly 4 men, we have two possibilities. 

Case 1.1. If the block of 4 men is at the end or the beginning of the row, we only need one women enclosing it. So we have the block $B$ consisting of 4 men and 1 woman, which can be at the beginning of the end. For the other $x-1$ women and block of two men, we have $x$ places. The block of men can stand anywhere, the $x-1$ women fill up the rest. 
Therefore this gives $2x$ possibilities.

Case 1.2. If the block of 4 men is not at the end or the beginning of the row, we need we need two women enclosing it. So we have the block $B$ consisting of 4 men and 2 woman. Further we have a block of 2 men. And then we have $x-2$ women. We therefore have $x$ places to place two unique objects, this can be done in $x(x-1)$ ways and again the women fill up the rest. 
This gives $x(x-1)$ possibilities. 

Case 2. Exactly 5 men is not possible because one men would be lonely.

Case 3. If there is a block of 6 men, and $x$ women, we have $x+1$ possibilities since there are $x+1$ places to place the block.

In total, we have $x^2+2x+1$ possibilities.
The total number is the way to order 3 blocks of two men on $x+3$ places, therefore it equals $$\frac{(x+3)(x+2)(x+1)}{6}$$
Plus the number of ways to order 2 blocks of three men on $x+2$ places, that equals $\frac{(x+2)(x+1)}{2}$.
Minus the number of ways to place one block a block of six men on $x+1$ places, that is $x+1$. 
Thus $$P=\frac{6(x^2+2x+1)}{(x^2+8x+6)(x+1)}=\frac{6(x+1)}{x^2+8x+6}=0.01$$
Solving this will give $x=594$. 
A: Consider the men as $M, M,\ldots,M$ to get 4 aligned and no men alone you need to form 2 groups, one with $4$ men and one with $2$ men. 
let $G(4),G(2)$ be the group with $4$ and  $2$ men respectively.
A favorable occurrence is of the form 
$$W(n_1) G(4) W(n_2) G(2) W(n_3) $$
$$W(n_1) G(2) W(n_2) G(4) W(n_3) $$
or $$ W(N_1) G(4) G(2) W(N_2)  $$
where $n_1 + n_2 +n_3 = N_1+N_2 = N$ the number of women.
Let's count how many favourable occurrences ($F$) there are. Since we are double counting ($G(4)G(2) = G(2)G(4)$) we get
$$F = {N+2 \choose 2} + {N+2 \choose 2} - {N+1 \choose 1} $$
The total of arrangements ($T$) when men are never apart is given by
the cases
$$ W(n_1)G(2)W(n_2)G(2) W(n_3)G(2) W(n_4) $$
or
$$ W(n_1)G(3)W(n_2)G(3) W(n_3)$$
 Since we are double counting ($G(2)G(2)G(2) = G(3)G(3)$) we get
$$ T = {N+3 \choose 3} +  {N+2 \choose 2} - {N+1 \choose 1}$$
Calculate the probability of success(p)
$$p = \frac{F}{T}  = \frac{{N+2 \choose 2} + {N+2 \choose 2} + {N+1 \choose 1}}{{N+3 \choose 3} +  {N+2 \choose 2} - {N+1 \choose 1}} $$
and check for which value of $N$ this is smaller than $1\%$.
$N = 594$
A: Let there be n women. 
To meet the criteria, men must either be in two blocks of 4 & 2 and positioned in two of the (n+1) gaps between women (including the ends), or in a block of 6 positioned in the (n+1) gaps,
i.e.  in $[(n+1)\cdot n  + (n+1)]$ patterns, having 4!2! and 6! permutations respectively, thus
$$\frac{[(n+1)\cdot n!]\cdot(n\cdot 4!\cdot2! + 6!)}{(6+n)!} \le \frac{1}{100},$$  which yields n = 7

EDIT
I took people as individuals, not as categories. Anyway, I am working out an answer treating as categories, and that in no arrangement can a man be left "lonely"
Basically, there will be (n+1) gaps (including ends) in which clumps of men can be positioned.
The favorable clumps are 4-2, 2-4 and 6.
The unfavorable clumps are 3-3 and 2-2-2
Thus $$\frac{2{n+1\choose 2} + {n+1\choose 1}}{2{n+1\choose 2} + {n+1\choose 1}  + {n+1\choose 2} + {n+1\choose 3}} \le \frac{1}{100}$$
Wolframalpha gives $\ge593.002$, thus 594
