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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?

Last updated date: 20th Jun 2024
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Answer
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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.