Question

Eight people decided to go for a picnic into two groups, one group to go by car and the other group by train. In how many ways can this be done if there must be at least 3 persons in each group?

Answer 1

We can consider the cases when both the group has the minimum number of persons. Then we can find the number of ways of selecting each group in each case and take their product to find the number of ways of grouping in each case. Then we can take their sum to find the total number of possible arrangements.

Complete step by step solution:
We are given that there are 8 persons and are divided into two groups. It is given that each group must have at least 3 members in it. So, we can consider different cases.
Case 1:
Consider the case where 3 persons go by car and rest 5 will go by train.
Out of 8, 3 people to go by car is selected in ${}^8{C_3}$ ways.
Then there will be 5 people left and they go by train. Its combination also can be written as ${}^5{C_5}$ .
Now the total number of ways of arranging the group in this case is given by taking the product.
$\Rightarrow {G_1} = {}^8{C_3} \times {}^5{C_5}$
We know that ${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$ . On expanding the combinations, we get,
$\Rightarrow {G_1} = \dfrac{{8!}}{{3!\left( {8 - 3} \right)!}} \times \dfrac{{5!}}{{5!\left( {5 - 5} \right)!}}$
On simplification we get,
$\Rightarrow {G_1} = \dfrac{{8 \times 7 \times 6 \times 5!}}{{3 \times 2 \times \left( 5 \right)!}} \times \dfrac{{5!}}{{5!\left( 0 \right)!}}$
On cancelling the common terms, we get,
$\Rightarrow {G_1} = 8 \times 7 \times 1$
On multiplication we get,
$\Rightarrow {G_1} = 56$
Case 2:
Now we can consider the case where 4 goes by car and 4 goes by train.
Out of 8, 4 people to go by car are selected in ${}^8{C_4}$ ways.
Then there will be 4 people left and they go by train. Its combination also can be written as ${}^4{C_4}$ .
Now the total number of ways of arranging the group in this case is given by taking the product.
$\Rightarrow {G_2} = {}^8{C_4} \times {}^4{C_4}$
We know that ${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$ . On expanding the combinations, we get,
$\Rightarrow {G_2} = \dfrac{{8!}}{{4!\left( {8 - 4} \right)!}} \times \dfrac{{4!}}{{4!\left( {4 - 4} \right)!}}$
On simplification we get,
$\Rightarrow {G_2} = \dfrac{{8 \times 7 \times 6 \times 5 \times 4!}}{{4 \times 3 \times 2 \times \left( 4 \right)!}} \times \dfrac{{4!}}{{4!\left( 0 \right)!}}$
On cancelling the common terms, we get,
$\Rightarrow {G_2} = 2 \times 7 \times 5 \times 1$
On multiplication we get,
$\Rightarrow {G_2} = 70$
Case 3:
Consider the case where 5 persons go by car and the rest 3 will go by train.
Out of 8, 5 people to go by car are selected in ${}^8{C_5}$ ways.
Then there will be 5 people left and they go by train. Its combination also can be written as ${}^3{C_3}$ .
Now the total number of ways of arranging the group in this case is given by taking the product.
$\Rightarrow {G_3} = {}^8{C_5} \times {}^3{C_3}$
We know that ${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$ . On expanding the combinations, we get,
$\Rightarrow {G_3} = \dfrac{{8!}}{{5!\left( {8 - 5} \right)!}} \times \dfrac{{3!}}{{3!\left( {3 - 3} \right)!}}$
On simplification we get,
$\Rightarrow {G_3} = \dfrac{{8 \times 7 \times 6 \times 5!}}{{3 \times 2 \times \left( 5 \right)!}} \times \dfrac{{3!}}{{3!\left( 0 \right)!}}$
On cancelling the common terms, we get,
$\Rightarrow {G_3} = 8 \times 7 \times 1$
On multiplication we get,
$\Rightarrow {G_3} = 56$
Case 4:
Now we can consider the case where 4 goes by train and 4 goes by car.
Out of 8, 4 people to go by train is selected in ${}^8{C_4}$ ways.
Then there will be 4 people left and they go by car. Its combination also can be written as ${}^4{C_4}$ .
Now the total number of ways of arranging the group in this case is given by taking the product.
$\Rightarrow {G_4} = {}^8{C_4} \times {}^4{C_4}$
We know that ${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$ . On expanding the combinations, we get,
$\Rightarrow {G_4} = \dfrac{{8!}}{{4!\left( {8 - 4} \right)!}} \times \dfrac{{4!}}{{4!\left( {4 - 4} \right)!}}$
On simplification we get,
$\Rightarrow {G_4} = \dfrac{{8 \times 7 \times 6 \times 5 \times 4!}}{{4 \times 3 \times 2 \times \left( 4 \right)!}} \times \dfrac{{4!}}{{4!\left( 0 \right)!}}$
On cancelling the common terms, we get,
$\Rightarrow {G_4} = 2 \times 7 \times 5 \times 1$
On multiplication we get,
$\Rightarrow {G_4} = 70$
So, the total number of ways of arrangement is given by the sum of all the 3 cases.
$\Rightarrow {G_1} + {G_2} + {G_3} + {G_4} = 56 + 70 + 56 + 70$
Hence on adding we get,
$\Rightarrow {G_1} + {G_2} + {G_3} + {G_4} = 252$

Therefore, 8 persons can be arranged into 2 groups such that each group has a minimum 3 persons in 252 ways.

Note:
We can consider only 3 cases as in all other cases one of the group will contain only less than 3 persons. While expanding the factorials, we need to expand till we get the biggest factorial in the denominator so that we can cancel them. We must find the product of the combinations not the sum to find the number of ways in individual cases. To find the total number of ways, we must take only the sum, not the product. We must note that $0! = 1$.