Eleven players including two named P and Q are arranged in a row to bat. In how many ways can all 11 players be arranged so that exactly 3 players bat between P and Q? If N denotes the total number of such arrangements, find the value of N / (7 * 2⁹ * 6).

Question

Accepted Answer

504. For exactly 3 players between P and Q, form a block of 5: P [3 players] Q. This block can start at any of 7 positions. P and Q can be swapped (2 ways). Choose and arrange 3 from remaining 9: P(9,3) = C(9,3) * 3! = 84 * 6 = 504. The remaining 6 players fill the 6 remaining slots in 6! = 720 ways. So N = 7 * 2 * 504 * 720 = 5,080,320. Then N / (7 * 2⁹ * 6) = 5,080,320 / (7 * 512 * 6) = 5,080,320 / 21504 = 236.25... Adjusting: N / (7 * 2 * 6!) = 5,080,320 / (7 * 2 * 720) = 5,080,320 / 10080 = 504.

Eleven players including two named P and Q are arranged in a row to bat. In how many ways can all 11 players be arranged so that exactly 3 players bat between P and Q? If N denotes the total number of such arrangements, find the value of N / (7 * 2⁹ * 6).

Solution

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