Question

there are 11 players in a cricket team from A to K find number of batting orders where A comes to bat before J ad J comes to bat after G

Answer 1

13305600

Answer 2

To determine the number of batting orders where A comes before J and J comes after G, we consider the relative positions of the three specific players: A, G, and J.

Let the total number of players be $N=11$. The total number of possible batting orders without any restrictions is $N! = 11!$.

Now, let's analyze the given conditions for players A, G, and J:

1. A comes to bat before J: This means that in any valid batting order, A must appear in a position earlier than J.
2. J comes to bat after G: This is equivalent to G comes to bat before J. This means G must appear in a position earlier than J.

Combining these two conditions, we know that J must bat after both A and G. However, there is no restriction on the relative order of A and G.

Let's consider the three specific players (A, G, J) and their possible relative arrangements. If we pick any three positions in the batting order, say $P_1, P_2, P_3$, these three players can be arranged in $3! = 6$ ways. These arrangements are:

1. (A, G, J)
2. (A, J, G)
3. (G, A, J)
4. (G, J, A)
5. (J, A, G)
6. (J, G, A)

Now, let's check which of these relative arrangements satisfy both conditions (A before J, and G before J):

• (A, G, J): A is before J, and G is before J. (Satisfies both conditions)
• (A, J, G): A is before J, but J is before G (G is not before J). (Does not satisfy condition 2)
• (G, A, J): G is before J, and A is before J. (Satisfies both conditions)
• (G, J, A): G is before J, but J is before A (A is not before J). (Does not satisfy condition 1)
• (J, A, G): J is before A and J is before G. (Does not satisfy either condition)
• (J, G, A): J is before A and J is before G. (Does not satisfy either condition)

Out of the 6 possible relative arrangements for A, G, and J, only 2 satisfy the given conditions: (A, G, J) and (G, A, J).

This means that for any set of three chosen positions for A, G, and J, there are 2 ways to arrange them such that the conditions are met, out of $3!$ total ways. Since all permutations are equally likely, the fraction of permutations that satisfy the conditions is $\frac{2}{3!}$.

Therefore, the total number of batting orders satisfying the conditions is: Total permutations $\times$ (Fraction of valid relative orders) $= 11! \times \frac{2}{3!}$ $= 11! \times \frac{2}{6}$ $= 11! \times \frac{1}{3}$

Now, we calculate the value: $11! = 39,916,800$

Number of batting orders $= \frac{39,916,800}{3} = 13,305,600$