We have $3$ guys and $2$ girls who are going to be seated on five lined-up chairs. The question is how many ways can they sit so that the girls stay next to each other.
My thought is that (if we number the chairs from 1 to 5) they either sit on 1-2 , 2-3 , 3-4 or 4-5, so $4$ ways for the girls. As for the guys, the first one has $3!$ ways, the second is $2!$, and the third is $1!$. So, the final number of ways is $4 \cdot 3! \cdot 2! \cdot 1!=48$ ways?
Best Answer
Your answer is correct, but your reasoning is not.
Method 1: There are four ways to place the leftmost girl. Since the girls are adjacent, choosing where the leftmost girl determines which two seats the girls occupy. There are $2!$ ways to arrange the two girls in those seats. The three boys can be arranged in the remaining three seats in $3!$ ways. Hence, there are $$4 \cdot 2!3! = 48$$ admissible arrangements.
Method 2: We can temporarily think of the girls as a single object. Then, we have four objects to arrange, the three boys and the block of girls. The objects can be arranged in $4!$ ways. The two girls can be arranged within the block in $2!$ ways. Hence, there are $$4!2! = 48$$ arrangements of three boys and two girls if the girls are adjacent.
What is wrong with your explanation?
Once you chose where the block of consecutive seats occupied by the girls begins, you have to arrange the two girls in the selected seats. Thus, there are $4 \cdot 2!$ ways to arrange the girls rather than $4$ ways. Also, the first boy may be seated in three ways (not $3!$, which is the number of ways of arranging three distinct objects in three positions), leaving two ways to seat the second boy, and one way to seat the third boy, giving $$4 \cdot 2! \cdot 3 \cdot 2 \cdot 1 = 4 \cdot 2! \cdot 3! = 4!2! = 48$$ admissible arrangements.