Explanation

(c)
Given, ASSISTANT$\rightarrow$AA I N SSS TT$\rightarrow$ STATISTICS$\rightarrow$AC II SSS TTT
Hence, N and C are not common
Same letters can be A, I, S, T
Probability of choosing ‘A’
$\frac{\ ^{2}C_{1}}{\ ^{9}C_{1}} \times \frac{\ ^{1}C_{1}}{\ ^{10}C_{1}} = \frac{2}{9} \times \frac{1}{10} = \frac{1}{45}$
Probability of choosing $‘I’$
$= \frac{1}{\ ^{9}C_{1}} \times \frac{\ ^{2}C_{1}}{\ ^{10}C_{1}} = \frac{1}{9} \times \frac{2}{10} = \frac{1}{45}$
Probability of choosing ‘$S$’
$= \frac{\ ^{3}C_{1}}{\ ^{9}C_{1}} \times \frac{\ ^{3}C_{1}}{\ ^{10}C_{1}} = \frac{3}{9} \times \frac{3}{10} = \frac{1}{10}$
Probability of choosing ‘$T$’
$= \frac{\ ^{2}C_{1}}{\ ^{9}C_{1}} \times \frac{\ ^{3}C_{1}}{\ ^{10}C_{1}} = \frac{2}{9} \times \frac{3}{10} = \frac{1}{15}$
Hence, required probability
$= \frac{1}{45} + \frac{1}{45} + \frac{1}{10} + \frac{1}{15} = \frac{19}{90}$