
In how many ways can you divide 12 persons into two groups of three people each and 3 groups of two people each?
A. 1663200
B. 4862200
C. 2796330
D. 5681030
Answer
584.4k+ views
Hint: We are given that there are a total of 12 people and we will be making 5 groups 3 of 2 people in one and two of 3 people in one. So from here we can see that the groups will somehow look like 2, 2, 2, 3, 3. We will use combinatorics to solve this question.
Complete step by step answer:
So this question is all about choosing the values here. First of all we will choose 2 people from the set of 12 people then 10 people remain, again we will choose 2 people from the rest 10. Now again we are left with 8 people, similarly choosing 2 more people from those 8 remaining and we will be left with 6 people. Now as we have 3 groups of 2 people now we will try to form 2 groups of 3 people. So now if we take 3 people from the remaining 6 to form a group we will be left with 3 people and those 3 people will be forming a group also, now it is very much clear that how the groups of 2, 2, 2, 3, 3 are being formed so let us do the same thing with combinatorics.
Choosing 2 people from the set of 12 will be given by \[{}^{12}{C_2}\]
Again choosing 2 from the remaining 10 will be given by \[{}^{10}{C_2}\]
And again choosing 2 from the remaining 8 will be \[{}^{8}{C_2}\]
Now as we are done with groups of 2 we will try to find combinations for groups of 3 which will be \[{}^{6}{C_3}\]
And the last group of 3 will be \[{}^{3}{C_3}\]
Now combining all of these we will be left with
\[\begin{array}{l}
\therefore {}^{12}{C_2} \times {}^{10}{C_2} \times {}^8{C_2} \times {}^6{C_3} \times {}^3{C_3}\\
= 66 \times 45 \times 28 \times 20 \times 1\\
= 1663200
\end{array}\]
So, the correct answer is “Option A”.
Note: Combinatorics is an area of mathematics primarily concerned with counting, both as a means and an end in obtaining results, and certain properties of finite structures. It is closely related to many other areas of mathematics and has many applications ranging from logic to statistical physics, from evolutionary biology to computer science, etc.
Complete step by step answer:
So this question is all about choosing the values here. First of all we will choose 2 people from the set of 12 people then 10 people remain, again we will choose 2 people from the rest 10. Now again we are left with 8 people, similarly choosing 2 more people from those 8 remaining and we will be left with 6 people. Now as we have 3 groups of 2 people now we will try to form 2 groups of 3 people. So now if we take 3 people from the remaining 6 to form a group we will be left with 3 people and those 3 people will be forming a group also, now it is very much clear that how the groups of 2, 2, 2, 3, 3 are being formed so let us do the same thing with combinatorics.
Choosing 2 people from the set of 12 will be given by \[{}^{12}{C_2}\]
Again choosing 2 from the remaining 10 will be given by \[{}^{10}{C_2}\]
And again choosing 2 from the remaining 8 will be \[{}^{8}{C_2}\]
Now as we are done with groups of 2 we will try to find combinations for groups of 3 which will be \[{}^{6}{C_3}\]
And the last group of 3 will be \[{}^{3}{C_3}\]
Now combining all of these we will be left with
\[\begin{array}{l}
\therefore {}^{12}{C_2} \times {}^{10}{C_2} \times {}^8{C_2} \times {}^6{C_3} \times {}^3{C_3}\\
= 66 \times 45 \times 28 \times 20 \times 1\\
= 1663200
\end{array}\]
So, the correct answer is “Option A”.
Note: Combinatorics is an area of mathematics primarily concerned with counting, both as a means and an end in obtaining results, and certain properties of finite structures. It is closely related to many other areas of mathematics and has many applications ranging from logic to statistical physics, from evolutionary biology to computer science, etc.
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