Question

A number consists of $7$ digits whose sum is $59$; prove that the chance of its being divisible by $11$ is $\frac{4}{21}$.

Answer 1

Let the sum of four numbers at odd places be $x$ and sum of three numbers at even places is $y$

Hence, $x+y=59$ and $x-y=11$

$\Rightarrow x=35$ and $y=24$

For formation of total sum $59$ with seven numbers following combinations are possible,

a) Number of permutations of ${}^{\prime }{9999995}^{\prime }=7$

b) Number of permutations of ${}^{\prime }{9999986}^{\prime }=42$

c) Number of permutations of ${}^{\prime }{9999977}^{\prime }=21$

d) Number of permutations of ${}^{\prime }{9999887}^{\prime }=105$

e) Number of permutations of ${}^{\prime }{9998888}^{\prime }=35$

Hence, total number of sample space $=7+42+21+105+35=210$

For formation of total sum $24$ with three numbers following combinations are possible,

a) Number of permutations of ${}^{\prime }{888}^{\prime }=1$

b) Number of permutations of ${}^{\prime }{789}^{\prime }=6$

c) Number of permutations of ${}^{\prime }{699}^{\prime }=3$

For sum $24$, total ways $=10$

For formation of total sum $35$ with four numbers following combinations are possible,

a) Number of permutations of ${}^{\prime }{9998}^{\prime }=4$

Hence, probability $=\frac{10\times 4}{210}=\frac{4}{21}$