Question

A box contains tickets numbered from 0 to 999999 . The probability of getting a ticket- with sum of digits on it being equal to 21 is given by $\frac{{ }^P C_5-6 \cdot{ }^q C_5+15 \cdot{ }^{ r } C_5}{10^6}$, find the value of $(p+q+r)$.

Answer 1

Concept: Coefficient if $x^r$ in $(1-x)^{-n}={ }^{n+r-1} C_r$
$n ( A )=$ coefficient of $x ^{21}$ in $\left(1+ x + x ^2+\ldots \ldots . .+ x ^9\right)^6$
(six digit starting with 000000,000001 etc. each place can have digit from 0 to 9)
$\therefore n ( A )=\text { coefficient of } x ^{21} \text { in }\left(\frac{1- x ^{10}}{1- x }\right)^6 $
$n ( A )=\text { coefficient of } x ^{21} \text { in }\left(1-{ }^6 C _1 x ^{10}+{ }^6 C _2 x ^{20}\right)(1- x )^{-6}$
$=\text { coefficient of } x^{21} \text { in }\left(1-{ }^6 x^{10}+15 x^{20}\right)(1-x)^{-6}$
$=\text { coefficient of } x^{21} \text { in }(1-x)^{-6}-6 \cdot \text { coefficient of } x^{11} \text { in }(1-x)^6+15 \text { coefficient of } x \text { in }(1-x)^{-6} $
$={ }^{26} C _{21}-6 \cdot{ }^{16} C _{11}+15 \cdot{ }^6 C _1 $
$={ }^{26} C _5-6 \cdot{ }^{16} C _5+15 \cdot{ }^6 C _5$
$\Rightarrow p + q + r =26+16+6=48$