Answer
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Hint:
Here, we are given a total of 5 people to be divided into 3 groups. We assume the two different cases i.e. (3, 1, 1) and (2, 2, 1), and with the help of combination formula, we can find the total number of possible ways.
Complete step-by-step answer:
Given, total number of people = 5.
Case I:
Either be put 3 persons in one group then two groups with 1 person,
Case II:
Two groups with 2 persons then one group of 1 person.
Total number of ways = \[{}^5{C_3} + {}^5{C_2} \times {}^3{C_2}\]
Total number of ways = \[{}^5{C_3} + {}^5{C_2} \times {}^3{C_2} \times \dfrac{1}{{2!}}\]
Here we divided the second part by 2! because two groups are there with 1 person.
⇒ Number of ways = $\dfrac{{5!}}{{3!2!}} + \dfrac{{5!}}{{2!3!}} \times \dfrac{{3!}}{{2!1!}} \times \dfrac{1}{{2!}}$
Here, 5! = 5 × 4 × 3 × 2 ×1 = 120
3! = 3 × 2 × 1 = 6
2! = 2 × 1 = 2
1! = 1
Putting all values, we get
Total number of ways = $\dfrac{{120}}{{6 \times 2}} + \dfrac{{120}}{{2 \times 6}} \times \dfrac{6}{2} \times \dfrac{1}{2} = 10 + \dfrac{{30}}{2} = 10 + 15 = 25$
Hence, the number of ways can five people be divided into three groups is 25.
Note:
In these types of questions always use the concept of the topic “Permutation and Combination”. But first we have to analyze whether the given problem is related to permutation or combination. Here we are required to divide the total number of peoples in groups, it does not matter how three people or two people are arranged in particular groups. So we have to use combination rules. Always keep in mind combination is used to divide the thing in particular group and permutation concepts is used for arrangement of things.
Here, we are given a total of 5 people to be divided into 3 groups. We assume the two different cases i.e. (3, 1, 1) and (2, 2, 1), and with the help of combination formula, we can find the total number of possible ways.
Complete step-by-step answer:
Given, total number of people = 5.
Case I:
Either be put 3 persons in one group then two groups with 1 person,
Case II:
Two groups with 2 persons then one group of 1 person.
Total number of ways = \[{}^5{C_3} + {}^5{C_2} \times {}^3{C_2}\]
Total number of ways = \[{}^5{C_3} + {}^5{C_2} \times {}^3{C_2} \times \dfrac{1}{{2!}}\]
Here we divided the second part by 2! because two groups are there with 1 person.
⇒ Number of ways = $\dfrac{{5!}}{{3!2!}} + \dfrac{{5!}}{{2!3!}} \times \dfrac{{3!}}{{2!1!}} \times \dfrac{1}{{2!}}$
Here, 5! = 5 × 4 × 3 × 2 ×1 = 120
3! = 3 × 2 × 1 = 6
2! = 2 × 1 = 2
1! = 1
Putting all values, we get
Total number of ways = $\dfrac{{120}}{{6 \times 2}} + \dfrac{{120}}{{2 \times 6}} \times \dfrac{6}{2} \times \dfrac{1}{2} = 10 + \dfrac{{30}}{2} = 10 + 15 = 25$
Hence, the number of ways can five people be divided into three groups is 25.
Note:
In these types of questions always use the concept of the topic “Permutation and Combination”. But first we have to analyze whether the given problem is related to permutation or combination. Here we are required to divide the total number of peoples in groups, it does not matter how three people or two people are arranged in particular groups. So we have to use combination rules. Always keep in mind combination is used to divide the thing in particular group and permutation concepts is used for arrangement of things.
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