Question

A four digit number (numbered from 0000 to 9999) is said to be lucky if sum of its first two digits is equal to the sum of its last two digits. If a four digit number is picked up at random, then the probability that it is a lucky number is:

• A. 0.07
• B. 0.067
• C. 0.67
• D. None

Answer 1

The correct option is B 0.067

Total number of ways of choosing a four digit number from (0000 to 9999) = ${10}^4$.
Let sum of first two digits and sum of last digits be k then number of ways will be $a_k^2$ where $a_k$ is number
of non – negative integral solutions of the equation $x_1+x_2=k,(0\le k\le 18)$
Then $a_k$ = Coefficienent of $x^k$ in
$(1+x+x^2+........+x^9{)}^2$
Then
$(1+x+x^2+........+x^9{)}^2$
$=a_0+a_{1}x+a_2{x}^2+........+a_{18}{x}^{18}.$
$a_0^2+a_1^2+........+a_{18}^2$ =constant term in
$(a_0+a_{1}x+.......+a_{18}{x}^{18})(a_0+\frac{a_1}{x}+........+\frac{a^{18}}{x^{18}})$
=const. term in $(1+x+.......+x^9{)}^2{(1+\frac{1}{x}+....+\frac{1}{x^9})}^2$
=Coefficient of $x^{18}$ in $(1+x+x^2+....+x^9{)}^4$
=Coefficient of $x^{18}$ in $(1-4x^{10})(1-x{)}^{-4}$
= coefficient of $x^{18}$ in $(1-4x^{10})(1-x{)}^{-4}$
${=}^{21}{C}_3-{4.}^{11}{C}_3=1330-660=670.$
Hence, required probability $=\frac{670}{10000}=0.067$