Each of the 7 children, in turn throws a ball once at a target. Calculate the number of ways the children can be arranged in order to take the throws.
Given that 3 of the children are girls and 4 are boys, calculate the number of ways the children can be arranged in order that
a. successive throws are made by boys and girls alternately,
b. a girl taken the first throw and a boy takes the last throw.
a. Well, let's see. We have 4 boys and 3 girls, and they need to take turns throwing the ball. So, the first throw can be taken by either a boy or a girl. Let's say we start with a boy. In that case, the second throw has to be taken by a girl. So we have 4 choices for the first boy, and then 3 choices for the first girl. After that, we have 3 boys and 2 girls left. Now, for the third throw, we need a boy again. That gives us 3 choices. And for the fourth throw, we need another girl, which gives us 2 choices. Finally, we have 2 boys and 1 girl left. So for the fifth throw, we have 2 choices, and for the last throw, we only have 1 choice left.
To calculate the total number of ways to arrange the children, we multiply all these choices together: 4 x 3 x 3 x 2 x 2 x 1 = 144.
b. Alright, let's do this! We have 4 boys and 3 girls, and we want a girl to take the first throw and a boy to take the last throw. So, the first throw has to be taken by one of the 3 girls. There are 3 choices. After that, we have 4 boys and 2 girls left. For the second throw, we can choose any one of the 4 boys, so we have 4 choices. Then, we have 3 boys and 2 girls left. For the third throw, we need a girl, so we have 2 choices. Now, we have 3 boys and 1 girl left. For the fourth throw, we need a boy, so we have 3 choices. Finally, we have 2 boys and 1 girl left, and the last throw has to be taken by a boy. So, we have 2 choices.
To calculate the total number of ways to arrange the children, we multiply all these choices together: 3 x 4 x 2 x 3 x 2 = 144.
So, for both cases, there are 144 ways to arrange the children. Shuffle them up and let the fun begin!
To calculate the number of ways the children can be arranged in order to take the throws, we can use the concept of permutations.
a. Successive throws made by boys and girls alternately:
In this case, we can think of arranging the 4 boys and 3 girls alternatively.
First, we arrange the 4 boys in a line, which can be done in 4! (4 factorial) ways. Similarly, we arrange the 3 girls in a line, which can be done in 3! ways.
Therefore, the total number of ways the children can be arranged in order to have successive throws made by boys and girls alternately is 4! * 3! = 24 * 6 = 144 ways.
b. A girl takes the first throw and a boy takes the last throw:
In this case, we need to choose one girl to take the first throw and one boy to take the last throw. Once these positions are fixed, we can arrange the remaining 3 boys and 2 girls in any order in between.
The number of ways to choose a girl for the first throw is 3, and the number of ways to choose a boy for the last throw is 4.
Once these positions are fixed, we can arrange the remaining 3 boys and 2 girls in any order in between. This can be done in (3+2)! = 5! = 120 ways.
Therefore, the total number of ways the children can be arranged in order to have a girl taking the first throw and a boy taking the last throw is 3 * 4 * 120 = 1440 ways.
To calculate the number of ways the children can be arranged in order to take the throws, we can use the concept of permutations.
For the first part, where we need to find the total number of arrangements without any particular condition, we can use the formula for permutations of a set with repeated elements:
n! / (n1! * n2! * ... * nk!)
where n is the total number of items and n1, n2, ..., nk are the number of repetitions of each item.
In our case, there are 7 children in total, with 3 girls and 4 boys. So, the total number of arrangements would be:
7! / (3! * 4!) = 7! / (3! * 2!) = 7 * 6 * 5 * 4! / (3! * 2!) = 7 * 6 * 5 = 210
So, there are 210 ways the children can be arranged in order to take the throws.
Now let's move on to the second part, where we have specific conditions.
a. To calculate the number of ways the children can be arranged in order, such that successive throws are made by boys and girls alternately, we can use similar logic.
In this case, we can consider the boys and girls as separate groups and arrange them within each group. The number of ways the boys can be arranged within their group is 4!, and the number of ways the girls can be arranged within their group is 3!. Since there are only two groups, the total number of arrangements would be the product of these two numbers:
4! * 3! = 24 * 6 = 144
So, there are 144 ways the children can be arranged in order, where successive throws are made by boys and girls alternately.
b. For the condition where a girl takes the first throw and a boy takes the last throw, we can again consider the boys and girls as separate groups and arrange them within each group.
In this case, we need to fix the positions of the first girl and the last boy. The number of ways the boys can be arranged within their group (excluding the last position) is 3!, and the number of ways the girls can be arranged within their group (excluding the first position) is 2!. The remaining positions can be filled by the remaining children, which is 4!.
So, the total number of arrangements would be:
2! * 3! * 4! = 2 * 6 * 24 = 288
Therefore, there are 288 ways the children can be arranged in order, where a girl takes the first throw and a boy takes the last throw.
no. of ways the children can be arranged to take the throws:
7p7 = 5040 ways.
for successive throws:
4p4 * 5p3 = 144 ways
for girl takes the first and boy the last:
3p1 * 5p5 *4p*1 = 1440 ways