
If we have two guys and three girls standing in line. Can they line up in as many different ways as possible and If the boys are on the ends, how many ways can they line up?
Answer
487.8k+ views
Hint: We begin addressing the problem by determining how the boys and girls can be organized in accordance with the problem's condition. The total number of ways to arrange two boys and three girls is then determined. We then find the ways the two boys can stand at the end and the number of ways 3 girls can arrange in the middle then multiply them.
Complete step-by-step solution:
We were given 2 guys and 3 girls standing in a line.
We will find how many ways they can stand in line.
We have a total number of students. The five youngsters are free to arrange themselves in whatever way they choose. As a result, we can choose one of the five children to be in position A. Then, in place B, one of the remaining four children. And so forth. As a result,
Total number of ways for arranging
$ = 5 \times 4 \times 3 \times 2 \times 1$
$ = 5!$
$ = 720$ ways
We will find the ways they can stand in a line if the boys are standing at the ends.
We can now look at how we may arrange the guys and the girls if we place the boys at different ends.
For placing boys - We may put them on the ends in two different ways, boys (boy 1 on the left or boy 2 on the left).
For arranging girls - We can arrange the girls in whatever manner we like in the middle, which is
$ = 3 \times 2 \times 1$
$ = 3!$
$ = 6$
So, the total number of ways they can arrange
$ = 2 \times 6$
$ = 12$ ways
Hence, Total number of ways we can arrange five students can arrange in line is 720 and if the girls are between them if we place the boys at different ends is 12.
Note: Whenever we have these sorts of issues, we strive to locate a suitable seating posture for both boys and girls, as this will provide us with the necessary information to solve the situation. We may also discover the total number of ways to arrange the boy by first selecting the locations for all two guys and then arranging them in those locations using combinations.
Complete step-by-step solution:
We were given 2 guys and 3 girls standing in a line.
We will find how many ways they can stand in line.
We have a total number of students. The five youngsters are free to arrange themselves in whatever way they choose. As a result, we can choose one of the five children to be in position A. Then, in place B, one of the remaining four children. And so forth. As a result,
Total number of ways for arranging
$ = 5 \times 4 \times 3 \times 2 \times 1$
$ = 5!$
$ = 720$ ways
We will find the ways they can stand in a line if the boys are standing at the ends.
We can now look at how we may arrange the guys and the girls if we place the boys at different ends.
For placing boys - We may put them on the ends in two different ways, boys (boy 1 on the left or boy 2 on the left).
For arranging girls - We can arrange the girls in whatever manner we like in the middle, which is
$ = 3 \times 2 \times 1$
$ = 3!$
$ = 6$
So, the total number of ways they can arrange
$ = 2 \times 6$
$ = 12$ ways
Hence, Total number of ways we can arrange five students can arrange in line is 720 and if the girls are between them if we place the boys at different ends is 12.
Note: Whenever we have these sorts of issues, we strive to locate a suitable seating posture for both boys and girls, as this will provide us with the necessary information to solve the situation. We may also discover the total number of ways to arrange the boy by first selecting the locations for all two guys and then arranging them in those locations using combinations.
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