Question

Rekha married Shivram and had 4 sons. Varsha married Ajoy and had 4 sons. Both the couples had divorce and after that Shivram married Varsha while Ajoy married Rekha. They too had 3 sons each from their wedlocks. How many selections of 8 children can be made from the 14 children so that each of them have an equal number of sons in the selection?

Answer 1

Hint: We solve for selection of children using a method of combination. Write initials of each couple as indication to the number of their sons. Count total number of children. Choose a pair of children from each collection of sons by different couples.
* Combination method helps us to find number of ways to choose r values from n values using the formula \[^n{C_r} = \dfrac{{n!}}{{(n - r)!r!}}\]where factorial of a number opens as \[n! = n.(n - 1).(n - 2)....3.2.1\]

Complete step-by-step answer:
We are given 4 persons Rekha, Shivram, Varsha nad Ajoy
We are given Rekha and Shivram had 4 sons, Varsha and Ajoy had 4 sons, Varsha and Shivram had 3 sons and Rekha and Ajoy had 3 sons.
Let us denote each of them by initials of their name, i.e.
Rekha-R; Shivram-S; Varsha-V and Ajoy-A
From each couple we write number of sons.
\[R + S = 4\] … (1)
\[V + A = 4\] … (2)
\[V + S = 3\] … (3)
\[R + A = 3\] … (4)
Total number of children\[ = (R + S) + (V + A) + (V + S) + (R + A)\]
\[ \Rightarrow \]Total number of children\[ = 4 + 4 + 3 + 3\]
\[ \Rightarrow \]Total number of children\[ = 14\] … (5)
We have to choose 8 children from 14 children in such a way that each person has equal number of sons in the selection.
If we choose 2 children from each set of children from equations (1), (2), (3) and (4) we will have total 8 children from 4 couples. In this way each of them will have equal number of children in the selection.
From equation (1):
We choose 2 children from 4 children.
Put \[n = 4,r = 2\]in formula of combination.
\[ \Rightarrow \]Number of ways to choose 2 children out of 4 children from equation (1) \[{ = ^4}{C_2}\]
Solve using the expansion of combination formula i.e.\[^n{C_r} = \dfrac{{n!}}{{(n - r)!r!}}\]
\[{ \Rightarrow ^4}{C_2} = \dfrac{{4!}}{{(4 - 2)!2!}}\]
\[{ \Rightarrow ^4}{C_2} = \dfrac{{4!}}{{2!2!}}\]
Open the factorial in the numerator as \[4! = 4 \times 3 \times 2!\]
\[{ \Rightarrow ^4}{C_2} = \dfrac{{4 \times 3 \times 2!}}{{2!2!}}\]
Cancel same terms from numerator and denominator.
\[{ \Rightarrow ^4}{C_2} = \dfrac{{4 \times 3}}{{2!}}\]
Write \[2! = 2 \times 1\]in the denominator.
\[{ \Rightarrow ^4}{C_2} = \dfrac{{4 \times 3}}{{2 \times 1}}\]
Cancel same terms from numerator and denominator.
\[{ \Rightarrow ^4}{C_2} = 2 \times 3\]
\[{ \Rightarrow ^4}{C_2} = 6\] … (6)
From equation (2):
We choose 2 children from 4 children.
Put \[n = 4,r = 2\]in formula of combination.
\[ \Rightarrow \]Number of ways to choose 2 children out of 4 children from equation (2) \[{ = ^4}{C_2}\]
Solve using the expansion of combination formula i.e.\[^n{C_r} = \dfrac{{n!}}{{(n - r)!r!}}\]
\[{ \Rightarrow ^4}{C_2} = \dfrac{{4!}}{{(4 - 2)!2!}}\]
\[{ \Rightarrow ^4}{C_2} = \dfrac{{4!}}{{2!2!}}\]
Open the factorial in the numerator as \[4! = 4 \times 3 \times 2!\]
\[{ \Rightarrow ^4}{C_2} = \dfrac{{4 \times 3 \times 2!}}{{2!2!}}\]
Cancel same terms from numerator and denominator.
\[{ \Rightarrow ^4}{C_2} = \dfrac{{4 \times 3}}{{2!}}\]
Write \[2! = 2 \times 1\]in the denominator.
\[{ \Rightarrow ^4}{C_2} = \dfrac{{4 \times 3}}{{2 \times 1}}\]
Cancel same terms from numerator and denominator.
\[{ \Rightarrow ^4}{C_2} = 2 \times 3\]
\[{ \Rightarrow ^4}{C_2} = 6\] … (7)
From equation (3):
We choose 2 children from 3 children.
Put \[n = 3,r = 2\]in formula of combination.
\[ \Rightarrow \]Number of ways to choose 2 children out of 3 children from equation (3) \[{ = ^3}{C_2}\]
Solve using the expansion of combination formula i.e.\[^n{C_r} = \dfrac{{n!}}{{(n - r)!r!}}\]
\[{ \Rightarrow ^3}{C_2} = \dfrac{{3!}}{{(3 - 2)!2!}}\]
\[{ \Rightarrow ^3}{C_2} = \dfrac{{3!}}{{1!2!}}\]
Open the factorial in the numerator as \[3! = 3 \times 2!\]
\[{ \Rightarrow ^3}{C_2} = \dfrac{{3 \times 2!}}{{1!2!}}\]
Cancel same terms from numerator and denominator.
\[{ \Rightarrow ^3}{C_2} = 3\] … (8)
From equation (3):
We choose 2 children from 3 children.
Put \[n = 3,r = 2\]in formula of combination.
\[ \Rightarrow \]Number of ways to choose 2 children out of 3 children from equation (4) \[{ = ^3}{C_2}\]
Solve using the expansion of combination formula i.e.\[^n{C_r} = \dfrac{{n!}}{{(n - r)!r!}}\]
\[{ \Rightarrow ^3}{C_2} = \dfrac{{3!}}{{(3 - 2)!2!}}\]
\[{ \Rightarrow ^3}{C_2} = \dfrac{{3!}}{{1!2!}}\]
Open the factorial in the numerator as \[3! = 3 \times 2!\]
\[{ \Rightarrow ^3}{C_2} = \dfrac{{3 \times 2!}}{{1!2!}}\]
Cancel the same terms from numerator and denominator.
\[{ \Rightarrow ^3}{C_2} = 3\] … (9)
Now selection of 8 children from 14 children is given by the product of equations (6), (7), (8) and (9)
Number of ways of selection \[{ = ^4}{C_2}{ \times ^4}{C_2}{ \times ^3}{C_2}{ \times ^3}{C_2}\]
Substitute the values from equations (6), (7), (8) and (9)
\[ \Rightarrow \]Number of ways of selection \[ = 6 \times 6 \times 3 \times 3\]
\[ \Rightarrow \]Number of ways of selection \[ = 324\]

Number of ways of selection of 8 children from 14 children is 324.

Note: Students might solve for the selection by substituting the value of n as 14 and r as 8 directly into the formula for combinations. This is the wrong process as the value obtained from this will be the number of ways of selection of 8 children from 14 children with no condition. So it will have all combinations and that may not have equal numbers of children for each person.