In how many way can 4 men and 3 women be arranged if 3 men always sit together?

I assume that you have four men, three women, each person is distinct, and we are wanting to count the number of ways that they can line up such that there are at least three men in a row (i.e. at least one man who has another man to both his left and his right).

I'm sure there are other ways to see the solution, but here I'll showcase one of them.

One of two things will happen, either all four men will be in a group together, or there will be at least one woman separating the men.

Arrange the women in a row ignoring the men for now ($3!$ options). Then, arrange the men in a row, ignoring the women ($4!$ options). Now, break into cases based on if there is going to be a group of all four men together or not and then interleave the men into slots between the women maintaining their order. If the men were all together, this could be done in $4$ ways, picking which slot between women they stand. If the men were split into a group of three and a group of one, then pick which slot was used for the group of three and then which is used for the group of one ($4$ options in the first case, $4\times 3$ options in the second case).

This gives us a total number of arrangements of $3!\times 4!\times (4 + 4\times 3)$

We could similarly have removed the "bad" arrangements where the men are all apart by splitting into cases similarly, but in this case it seems like more work rather than less. It would give $7! - 4!3!(\binom{4}{4} + 4\times \binom{3}{2} + \binom{4}{2})$ which should come to the same value as before.


Start by picking the group of three men who sit together. This can be done in $\binom{4}{3}$ ways. Consider this group as one object. We then have five objects to arrange: the group of three, the remaining man, and the three women. The men in the group of three can be arranged in $3!$ ways. So in all, we have $\binom{4}{3}5!3!$ possible arrangements.

But there is a problem. If there is a group of four men, say abcd, that sit together then we have counted each such group twice, once as a(bcd) and once as (abc)d. So we need to count these arrangements and subtract them from our first count. Consider the four men as one object. We then have four objects to arrange: the group of men, and the three women. The men in the group can be arranged in $4!$ ways. So there are $4!4!$ arrangements.

The final answer is $$\binom{4}{3}5!3! - 4!4!$$

Tags:

Combinatorics

Related

Almost Sure and Mean Square Convergence (Check My Answer) Is every prime contained in the set $\,\{n\in\mathbb{N}:n|10^k-1\}\cup\{2,5\}\,$ for $k=1,2,3,...$? Is there a sequence $x_n\to+\infty$ such that $\liminf x_{2n}/x_n = 0$? Integral representation of the Euler-Mascheroni constant involving $\pi$ Compute $\int_{\mathbb R}\frac{dx}{1+2x^2+x^4}$ $ \int_0^\frac{\pi}{2}\ln^n\left(\tan(x)\right)\:dx$ It is impossible for $(x-1)^2+x^2+(x+1)^2$ to be a perfect square Express $\sum_{j=1}^{n}\sum_{i=1}^{n} \frac{1}{i(i+j)}$ in terms of harmonic numbers Every interval $I \subset \mathbb{R}$ is connected. [Proof clarification] Is this a scalar product of three vectors? Has $n^{n+1}+(n+1)^{n+2}$ other obvious factors than that I found? Can I get a sequence of bounded functions converging pointwise to $f(x)=1/x$ for $x$ non zero and $0$ for $x$ zero?

Recent Posts