Question

The total number of ways of distributing $21$ books among $5$ children

• A. without any restriction is ${}^{25}{C}_4$
• B. when each child should get atleast one book is ${}^{16}{C}_4$
• C. when each child should get atleast three books is ${}^{10}{C}_4$
• D. when each child should get atleast four books is $5$

Answer 1

Answer: (A), (C), (D)

The correct options are
A without any restriction is ${}^{25}{C}_4$
C when each child should get atleast three books is ${}^{10}{C}_4$
D when each child should get atleast four books is $5$

Let the number of books recieved by children be $x_1,x_2,x_3,x_4,x_5$
$\left(i\right)$Here, $x_1+x_2+x_3+x_4+x_5=21$
Total ways of distribution without any restriction is $={}^{21+5-1}{C}_{5-1}={}^{25}{C}_4$

$\left(ii\right)$Now, $x_1+x_2+x_3+x_4+x_5=21,x_1,x_2,x_3,x_4,x_5\ge 1$
Let $X_1=x_1-1,X_2=x_2-1,X_3=x_3-1,X_4=x_4-1,X_5=x_5-1$
So, $X_1+X_2+X_3+X_4+X_5=21-5=16,X_1,X_2,X_3,X_4,X_5\ge 0$
Total ways of distribution when each child should get atleast one book is $={}^{16+5-1}{C}_{5-1}={}^{20}{C}_4$


$\left(iii\right)$Now, $x_1+x_2+x_3+x_4+x_5=21,x_1,x_2,x_3,x_4,x_5\ge 3$
Let $X_1=x_1-3,X_2=x_2-3,X_3=x_3-3,X_4=x_4-3,X_5=x_5-3$
So, $X_1+X_2+X_3+X_4+X_5=21-15=6,X_1,X_2,X_3,X_4,X_5\ge 0$
Total ways of distribution when each child should get atleast one book is $={}^{6+5-1}{C}_{5-1}={}^{10}{C}_4$


$\left(iv\right)$Now, $x_1+x_2+x_3+x_4+x_5=21,x_1,x_2,x_3,x_4,x_5\ge 4$
Let $X_1=x_1-4,X_2=x_2-4,X_3=x_3-4,X_4=x_4-4,X_5=x_5-4$
So, $X_1+X_2+X_3+X_4+X_5=21-20=1,X_1,X_2,X_3,X_4,X_5\ge 0$
Total ways of distribution when each child should get atleast one book is $={}^{1+5-1}{C}_{5-1}={}^5{C}_4=5$