In how many ways the word 'PERMUTATIONS' can be arranged such that 4 alphabets exist between the alphabet P and S?
I don't have an answer to this question.
Following is the approach that I followed. Is it correct or any other better approach to solve this problem?
Number of ways P and S can be placed such that 4 alphabets are present between them = 8 i.e. (1,5) (2,6) (3,7) (4,8) (5,9) (6,10) (7,11) (8,12)
Similarly, number of ways S and P can be placed such that 4 alphabets are present between them = 8
Total number of ways = 8 * 2 = 16
Remaining 10 positions can be occupied in 10 P(10) = 10!
Total permutations = 16 * 10!
I don't have an answer to this question.
Following is the approach that I followed. Is it correct or any other better approach to solve this problem?
Number of ways P and S can be placed such that 4 alphabets are present between them = 8 i.e. (1,5) (2,6) (3,7) (4,8) (5,9) (6,10) (7,11) (8,12)
Similarly, number of ways S and P can be placed such that 4 alphabets are present between them = 8
Total number of ways = 8 * 2 = 16
Remaining 10 positions can be occupied in 10 P(10) = 10!
Total permutations = 16 * 10!
