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Here P and S are fixed, so 10 letters are left, out of which 2 are T's

So, 10 letters out of which 2 are T's can be arranged in 10!/2! = 1814400 ways

Now letters P and S can be arranged so that there are 4 letters between them, which can be done in 2*7 = 14 ways

Required no. of ways = 1814400 * 14 = 25401600

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Q: In how many ways can the letters of the word permutations be arranged if there are always 4 letters between p and s?

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You can simplify the problem by considering it as two different problems. The first involves consider the five-book chunk as a single book, and calculating the permutations there. The second involves the permutations of the books within the five-book block. Multiplying these together gives you the total permutations. Permutations of five objects is 5!, five gives 5!, so the total permutations are: 5!5! = 5*5*4*4*3*3*2*2 = 263252 = 14,400 permutations

The different ways the vowels can be arranged together is 3x2x1 = 6. The number of places this arrangement can go within the word is 5. This means there are 30 possible arrangements for the vowels. The consonants can be arranged amongst themselves with 4 possibilities for first place, 3 for second, 2 for third and 1 for fourth. This gives us 4x3x2=24 combinations. So with 30 possible arrangements for the vowels, and 24 for the consonants, this gives us 720 possible ways of arranging the letters in the word offices so that the vowels always come together.

Yes

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The range of a distribution is the difference between the largest number in the range and the smaller number. ?æThe number will always be positive, but may not always be whole.

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No. The number of permutations or combinations of 0 objects out of n is always 1. The number of permutations or combinations of 1 object out of n is always n. Otherwise, yes.

You can simplify the problem by considering it as two different problems. The first involves consider the five-book chunk as a single book, and calculating the permutations there. The second involves the permutations of the books within the five-book block. Multiplying these together gives you the total permutations. Permutations of five objects is 5!, five gives 5!, so the total permutations are: 5!5! = 5*5*4*4*3*3*2*2 = 263252 = 14,400 permutations

the word leading contains 7 different letters,when the vowels eai r always together they can be supposed to form 1 letter then we have to arrange ldng (eai) now,5 letters can be arranged in 5!ways=120 ways the vowels (eai)can be arranged in 3! ways=6ways required no of ways =120*6=720 by deepika.R(RVS CET)

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State run lotteries are prime examples of businesses that use permutations and combinations to design plays that will always stack the odds against the players and assure the states with a tidy profit be it for education or other expenses

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