Question

The greatest common divisor of \[{}^{20}{C_1},{}^{20}{C_3},...,{}^{20}{C_9}\] is
A. \[20\]
B. \[4\]
C. \[5\]
D. None of these

Answer 1

Hint: In the given question, we need to find the greatest common divisor of \[{}^{20}{C_1},{}^{20}{C_3},...,{}^{20}{C_9}\] . For this, we will use the formula of combination to find the values of \[{}^{20}{C_1},{}^{20}{C_3},{}^{20}{C_5},{}^{20}{C_7},{}^{20}{C_9}\] . After that, we will find the greatest common divisor of it using their prime factors.

Formula used: The following formula used for solving the given question.
The formula of combination is given by, \[{}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}\]
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: We know that \[{}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}\]
Here, \[n\] is the total number of things and \[r\] is the number of things that need to be selected from total things.
Now, we will find the values of \[{}^{20}{C_1},{}^{20}{C_3},{}^{20}{C_5},{}^{20}{C_7},{}^{20}{C_9}\].
 \[{}^{20}{C_1} = \dfrac{{20!}}{{1!\left( {20 - 1} \right)!}}\]
By simplifying, we get
\[{}^{20}{C_1} = 20\]
This gives \[{}^{20}{C_1} = 2 \times 2 \times 5\]
Now, we have \[{}^{20}{C_3} = \dfrac{{20!}}{{3!\left( {20 - 3} \right)!}}\]
By simplifying, we get
\[{}^{20}{C_3} = \dfrac{{20!}}{{3!17!}}\]
\[{}^{20}{C_3} = 1140\]
So, we get \[{}^{20}{C_3} = 2 \times 2 \times 3 \times 5 \times 19\]
Let us find the value of \[{}^{20}{C_5} = \dfrac{{20!}}{{5!\left( {20 - 5} \right)!}}\]
By simplifying, we get
\[{}^{20}{C_5} = \dfrac{{20!}}{{5!\left( {15} \right)!}}\]
\[{}^{20}{C_5} = 15504\]
Thus, we get \[{}^{20}{C_5} = 2 \times 2 \times 2 \times 2 \times 3 \times 17 \times 19\]
Now, we will find the value of \[{}^{20}{C_7} = \dfrac{{20!}}{{7!\left( {20 - 7} \right)!}}\]
By simplifying, we get
\[{}^{20}{C_7} = \dfrac{{20!}}{{7!13!}}\]
\[{}^{20}{C_7} = 77520\]
This gives, \[{}^{20}{C_7} = 2 \times 2 \times 2 \times 2 \times 3 \times 5 \times 17 \times 19\]
Lastly, we will have \[{}^{20}{C_9} = \dfrac{{20!}}{{9!\left( {20 - 9} \right)!}}\]
By simplifying, we get
\[{}^{20}{C_9} = \dfrac{{20!}}{{9!\left( {11} \right)!}}\]
\[{}^{20}{C_9} = 167960\]
This gives, \[{}^{20}{C_9} = 2 \times 2 \times 2 \times 5 \times 13 \times 17 \times 19\]
Hence, from the factors of the above numbers, the greatest common factor is \[(2 \times 2)\].
The greatest common factor is \[4\] .

Thus, Option (B) is correct.

Additional Information : Combinations are ways to choose elements from a group in mathematics in which the order of the selection is irrelevant. Let's say we have a trio of numbers. Then, combination determines how many ways we can choose two numbers from each group.

Note: Many students make mistakes in the calculation part as well as writing the combination rule. This is the only way through which we can solve the example in the simplest way. Also, it is essential to find correct prime factorization of given numbers to get the desired result.