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

Answer: 48

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