Combinatorics: pick 2 from group A with 5 elements and 3 that stay in order from group B with 6 elements?

Question

So basically: how many way are there to pick 2 elements from group A which has 5 elements, and pick 3 elements from group B which has 6 elements, provided the 3 elements stay in order in the new 5 element picked set?

Answer 1

So we start by choosing $2$ elements from $A$ in $\binom{5}{2}$ ways. Then we permute three elements of $B$, which can be done in $\frac{6!}{3!}$. We then cluster these three elements and permute the three clusters of our new set in $3!$ ways. By rule of product, we multiply:

$$\binom{5}{2} \cdot \frac{6!}{3!} \cdot 3!$$

If $B$ is already ordered and we wish to preserve that order, we have to do things differently. So we start by simply choose $3$ elements of $B$, which fixes their order. So we do this in $\binom{5}{3}$ ways. That leaves us with:

$\_b_{1} \_ b_{2} \_ b_{3} \_$

Now we permute the $a_{1}, a_{2}$ elements into the String (filling in the blanks) in $\frac{4!}{2!}$ ways. So by rule of product, we multiply:

$$\binom{6}{3} \cdot \binom{5}{2} \cdot \frac{4!}{2!}$$

Edit: Based on your comment, I will update my solution.

So we start by permuting the three girls, which can be done in $6!/3!$ ways. We permute because this is a line, so order matters. We now choose our two boys in $\binom{5}{2}$ ways. Now we treat the cluster of girls as an object and each boy as an object. We have three objects to permute, so we multiply by $3!$ to get: $6! \cdot \binom{5}{2}$ ways of arranging everyone in a line.

Answer 2

If I understand you correctly, we are looking at the following situation:

• The set $A$ has $5$ elements.
• The set $B$ has $6$ elements.
• You want to arrange $2$ elements from $A$ and $3$ elements from $B$ in a line, such that the elements of $B$ are adjacent.

Now, the general arrangements you're looking at are of one of the following three forms:

1. $AABBB$
2. $ABBBA$
3. $BBBAA$

Pick any one of the three forms. How many ways are there to arrange $3$ of the $6$ elements of $B$ in the $B$ slots? How many ways are there to arrange $2$ of the elements of $A$ in the $A$ slots? Can you take it from there, given that there are $3$ forms? Feel free to run any answer attempts by me, and I'll check them.