Question

There are \[10\] bags \[{B_1},{B_2},{B_3},{B_4},...{B_{10}}\] 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. \[{}^{32}{C_{20}}\]

Answer 1

Hint: In the given question, we need to find the total number of ways to bring out \[10\] articles from a bag. For this, we will find first the expression of the total number of ways using the combination formula and we will simplify the expression. After that, we will use the following formula to get the desired result.

Formula used: The following formula used for solving the given question.
The total possibilities is given by, \[{}^n{C_r} + {}^n{C_{r - 1}} = {}^{n + 1}{C_r}\]
Here, \[n\] is the total number of things and \[r\] is the number of things that need to be selected from total things.

Complete step by step solution: The expression for the total number of ways is given by
 \[{}^{21}{C_{10}} + {}^{22}{C_{10}} + ..... + {}^{30}{C_{10}} = {}^{21}{C_{11}} + {}^{22}{C_{12}} + {}^{23}{C_{13}} + .... + {}^{30}{C_{20}}\]
Thus by adding and subtracting \[{}^{21}{C_{10}}\] on both sides, we get
\[{}^{21}{C_{10}} + {}^{22}{C_{10}} + ..... + {}^{30}{C_{10}} + {}^{21}{C_{10}} - {}^{21}{C_{10}} = {}^{21}{C_{11}} + {}^{22}{C_{12}} + {}^{23}{C_{13}} + .... + {}^{30}{C_{20}} + {}^{21}{C_{10}} - {}^{21}{C_{10}}\]By simplifying, we get
\[({}^{21}{C_{11}} + {}^{21}{C_{10}}) + {}^{22}{C_{12}} + {}^{23}{C_{13}} + {}^{24}{C_{14}} + {}^{25}{C_{15}} + {}^{26}{C_{16}} + {}^{27}{C_{17}} + {}^{28}{C_{18}} + {}^{29}{C_{19}} + {}^{29}{C_{20}} - {}^{21}{C_{10}}\]
By using the formula \[{}^n{C_r} + {}^n{C_{r - 1}} = {}^{n + 1}{C_r}\], we get
\[({}^{22}{C_{11}} + {}^{22}{C_{12}}) + {}^{23}{C_{13}} + {}^{24}{C_{14}} + {}^{24}{C_{14}} + {}^{25}{C_{15}} + {}^{26}{C_{16}} + {}^{27}{C_{17}} + {}^{28}{C_{18}} + {}^{29}{C_{19}} + {}^{30}{C_{20}} - {}^{21}{C_{10}}\]
Hence, by simplifying further we get
\[({}^{23}{C_{12}} + {}^{23}{C_{13}}) + {}^{24}{C_{14}} + {}^{25}{C_{15}} + {}^{26}{C_{16}} + {}^{27}{C_{17}} + {}^{28}{C_{18}} + {}^{29}{C_{19}} + {}^{30}{C_{20}} - {}^{21}{C_{10}}\]
By simplifying this way, finally we get
\[{}^{31}{C_{20}} - {}^{21}{C_{10}}\]

Thus, Option (A) is correct.

Additional Information: Probability refers to possibility. A random event's occurrence is the subject of this area of mathematics. The range of the value is zero to one. Mathematics has incorporated probability to forecast the likelihood of various events.

Note: Many students make mistakes in the calculation part as well as writing the total possibility rule. This is the only way through which we can solve the example in the simplest way. Also, it is essential to analyze the result of \[{}^n{C_r} + {}^n{C_{r - 1}} = {}^{n + 1}{C_r}\] to get the desired result.