Question

In how many ways a team of 10 players out of 22 players can be made if 6 particular players are always to be included and 4 particular players are always excluded
$A){}^{22}{C_{10}}$
$B){}^{18}{C_3}$
$C){}^{12}{C_4}$
$D){}^{18}{C_4}$

Answer 1

Hint: To solve this question, we should know the details about the non-fixed places in the team and the players who are always excluded or included. The number of ways of selecting “r” players out of players is given by${}^n{C_r}$. Using this formula, we can get the answer.

Complete step by step answer:
In the question, It is given that a team of $10$ players out of $22$ players. Here are $6$ particular players always to be included and $4$ particular players are always excluded.
Therefore, we can write the total numbers of players $ = 22$
We have to find that we need to select the team of $10$ players
We have to exclude $4$ particulars players of them
So we have to subtract it as the total number of players
Now we have only $18$ players are now available.
Also, from these $6$ particulars will always be including.

$\therefore$ The required number of ways $ = {}^{12}{C_4}$.

Note:
In these types of problems, we need to know the key concept of permutation and combination. Students can make a mistake by not considering the constraints given in the question.
That leads to a selection of $10$ out of $22$ players which can be written as ${}^{22}{C_{10}}$$ = {}^{22}{C_{22 - 10}} = {}^{22}{C_{12}}$ which leads to a wrong answer.
Generally, students get confused between permutation and combination.
If you have to select, then use a combination, and if you have to arrange use permutation. It is a very nice trick to use. Do not forget to use the correct way otherwise you will get the wrong answer.