Let 10 Bags $B_1, B_2, \ldots, B_{10}$ which contains $21, 22, \ldots, 30$ different articles respectively. Then the total number of ways to bring out 10 articles from a Bag is

The total number of ways to bring out 10 articles from a bag is the sum of the number of ways to choose 10 articles from each of the 10 bags.

Number of ways to choose 10 articles from bag $B_1$ containing 21 articles is $^{21}C_{10}$.

Number of ways to choose 10 articles from bag $B_2$ containing 22 articles is $^{22}C_{10}$.

Similarly, the number of ways for bag $B_{10}$ containing 30 articles is $^{30}C_{10}$.

Total number of ways $S$ is given by:

$S = ^{21}C_{10} + ^{22}C_{10} + ^{23}C_{10} + \ldots + ^{30}C_{10}$

Adding and subtracting $^{21}C_{11}$ to the series:

$S = (^{21}C_{11} + ^{21}C_{10}) + ^{22}C_{10} + \ldots + ^{30}C_{10} - ^{21}C_{11}$

Using the identity $^{n}C_{r} + ^{n}C_{r-1} = ^{n+1}C_{r}$:

$S = ^{22}C_{11} + ^{22}C_{10} + \ldots + ^{30}C_{10} - ^{21}C_{11}$

$S = ^{23}C_{11} + ^{23}C_{10} + \ldots + ^{30}C_{10} - ^{21}C_{11}$

Continuing this process by combining adjacent terms:

$S = ^{30}C_{11} + ^{30}C_{10} - ^{21}C_{11}$

$S = ^{31}C_{11} - ^{21}C_{11}$

Using the property $^{n}C_{r} = ^{n}C_{n-r}$:

$^{31}C_{11} = ^{31}C_{20}$ and $^{21}C_{11} = ^{21}C_{10}$

Therefore, $S = ^{31}C_{20} - ^{21}C_{10}$

Answer: ${}^{31}C_{20} - {}^{21}C_{10}$