Question

In a cricket match against Pakistan, Azhar wants to bat before Jadeja and Jadeja wants to bat before Ganguli. The number of possible batting orders with the above restrictions, if the remaining eight team members are prepared to bat at any given place, is
A) $\dfrac{{11!}}{{3!}}$
B) ${}^{11}{P_3}$
C) $\dfrac{{11!}}{3}$
D) None

Answer 1

Hint:: We should have the concept of permutation to solve this problem. In the question, we have to find out the number of ways the players will play in order along with given restrictions. Apply the concept of permutation according to given restrictions find out the total no. of ways to arrange all players fixing the positions of $3$ players. For solving this, the factorial of total no. of players ($11$ in a cricket team) should be multiplied by the factorial of no. of players having no restriction & this should be divided by-product of the factorial of no. of players having restriction (3 here) & factorial of no. of players having no restriction (8 here).

Complete step by step solution:
The total number of players in a cricket match is $ = 11$players.
Now, given restrictions are – Azhar, Jadeja, and Ganguli.
Since it is given that Azhar will play before Jadeja and Jadeja will play before Ganguli.
According to question Azhar, Jadeja and Ganguli can play in any order\[like{\text{ }}1,2,3{\text{ }}or{\text{ }}2,3,4{\text{ }}or{\text{ }}3,4,5{\text{ }}etc\] i.e. Azhar can take any position from \[1{\text{ }}to{\text{ }}9\] but should be one after another (Azhar then Jadeja then Ganguli ) as mentioned in question.
So, their order or places are fixed.
Remaining players in team $ = (11 - 3) = 8$players.
$\Rightarrow$ So, they will arrange in $ = 8!$ways and fixed order are $ = 3$.
$\Rightarrow$ $\dfrac{{Total{\text{ }}no.{\text{ }}of{\text{ }}players!}}{{no.of{\text{ }}players{\text{ with }}restriction!{\text{ }} \times {\text{ }}no.{\text{ }}of{\text{ }}players{\text{ with }}no{\text{ }}restrictions!}}$\[ \times {\text{ }}no.{\text{ }}of{\text{ }}players{\text{ }}having{\text{ }}no{\text{ }}restrictions\]
$ = \dfrac{{11!}}{{3! \times 8!}} \times 8!$
On simplifying the above terms we get,
$ = \dfrac{{11!}}{{3!}}$

$\therefore$ No. of possible batting order $ = \dfrac{{11!}}{{3!}}$. Hence option (A) is the correct answer.

Note: First of all, read the question properly to understand given restrictions in the question so that you can relate the way needed to solve the sum in your mind. The above question was asked from permutation. It is a conceptual topic and required great concentration specially on the restrictions given in the question. Sometimes there is a chance of using the wrong formula for solving because the formula for this type of problem depends completely on what restriction is given regarding arrangements so a crystal-clear concept is required to solve these questions.