Find the number of ways in which we can select three numbers from 1 to \[30\] so as to exclude every selection of all even numbers.
A. \[4060\]
B. \[3605\]
C. \[455\]
D. None of these

Hint: First, calculate the number of ways to select the 3 numbers from 1 to \[30\]. Then, find the number of ways to select the 3 even numbers from 1 to \[30\]. In the end, subtract the total number of ways to select the 3 even numbers from 1 to \[30\] from the total number of ways to select the 3 numbers from 1 to \[30\] and solve it to get the required answer.



Formula Used:The combination formula: \[{}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}\]
Factorial: \[n! = 1 \times 2 \times ... \times n = n\left( {n - 1} \right)!\]



Complete step by step solution:The given numbers are \[1,{\rm{ }}2,{\rm{ }}3,{\rm{ }}4,{\rm{ }}....,{\rm{ }}30\].
The total number of ways to select the 3 numbers from 1 to \[30\]: \[{}^{30}{C_3}\] \[.....\left( 1 \right)\]

Now find the number of ways to select the 3 even numbers from 1 to \[30\].
We know that, \[2,{\rm{ 4}},{\rm{ 6}},{\rm{ }}....,{\rm{ }}30\] are the \[15\] even numbers.
Therefore, the number of ways to select the 3 even numbers from 1 to \[30\]: \[{}^{15}{C_3}\] \[.....\left( 2 \right)\]

To calculate the number of ways in which we can select three numbers from 1 to \[30\] so as to exclude every selection of all even numbers, subtract equation \[\left( 2 \right)\] from the equation \[\left( 1 \right)\].
We get,
\[{}^{30}{C_3} - {}^{15}{C_3} = \dfrac{{30!}}{{3!\left( {30 - 3} \right)!}} - \dfrac{{15!}}{{3!\left( {15 - 3} \right)!}}\]
\[ \Rightarrow {}^{30}{C_3} - {}^{15}{C_3} = \dfrac{{30!}}{{3!27!}} - \dfrac{{15!}}{{3!12!}}\]
Simplify the right-hand side by applying the formula \[n! = 1 \times 2 \times ... \times n = n\left( {n - 1} \right)!\]
\[ \Rightarrow {}^{30}{C_3} - {}^{15}{C_3} = \dfrac{{30 \times 29 \times 28 \times 27!}}{{3!27!}} - \dfrac{{15 \times 14 \times 13 \times 12!}}{{3!12!}}\]
\[ \Rightarrow {}^{30}{C_3} - {}^{15}{C_3} = \dfrac{{30 \times 29 \times 28}}{6} - \dfrac{{15 \times 14 \times 13}}{6}\]
\[ \Rightarrow {}^{30}{C_3} - {}^{15}{C_3} = 4060 - 455\]
\[ \Rightarrow {}^{30}{C_3} - {}^{15}{C_3} = 3605\]
Thus, the number of ways in which we can select three numbers from 1 to \[30\] so as to exclude every selection of all even numbers is \[3605\].



Option ‘B’ is correct



Note: Students often get confused and consider the combination formula as \[{}^n{C_r} = \dfrac{{n!}}{{\left( {n - r} \right)!}}\], which is an incorrect formula. Because of this, they get the wrong solution.