Question
There are 10 bags $${B_1},{B_2},{B_3},.....,{B_{10}},$$ which contain 21,22, . . . . . , 30 different articles respectively. The total number of ways to bring out 10 articles from a bag is
A.
$$^{31}{C_{20}} - {\,^{21}}{C_{10}}$$
B.
$$^{31}{C_{21}}$$
C.
$$^{31}{C_{20}}$$
D.
None of these
Answer :
$$^{31}{C_{20}} - {\,^{21}}{C_{10}}$$
Solution :
The required number of ways $${ = ^{21}}{C_{10}} + {\,^{22}}{C_{10}} + {\,^{23}}{C_{10}} + ..... + {\,^{30}}{C_{10}}$$
$$\eqalign{
& = \left( {^{21}{C_{10}} + {\,^{21}}{C_{11}} + {\,^{22}}{C_{12}} + {\,^{23}}{C_{13}} + ..... + {\,^{30}}{C_{20}}} \right) - {\,^{21}}{C_{10}}\left( {\because \,{\,^n}{C_r} = {\,^n}{C_{n - r}}} \right) \cr
& = {\,^{22}}{C_{11}} + {\,^{22}}{C_{12}} + {\,^{23}}{C_{13}} + ..... + {\,^{30}}{C_{20}} - {\,^{21}}{C_{10}}\left( {\because \,{\,^n}{C_{r - 1}} + {\,^n}{C_r} = {\,^{n + 1}}{C_r}} \right) \cr
& ......................................................................... \cr
& = {\,^{31}}{C_{20}} - {\,^{21}}{C_{10}}. \cr} $$