Letters from the word "MISSISSIPPI" are used to form words. Each letter can be used at most once.?

Question

a)How many 11-letter words can be formed?
i)How many of those starting with p?
ii)How many of those having all S's together?

b) How many 10-letter words can be formed?
i) How many of those having all four I's together?

Answer 1

a) Mississippi is an 11-letter word, so any 11-letter word formed from it is simply a permutation of the letters. There are 11! such permutations, but not all of them are distinct, because some of the letters are identical. The number of times each distinct word will be repeated in a list that assumes that the Ss, Ps, and Is are all distinct is simply the number of ways the Ss can be permuted amongst themselves times the number of ways the Ps can be permuted amongst themselves times the number of ways the Is can be permuted amongst themselves, which is 4!*2!*4!. Therefore, the total number of distinct permutations is 11!/(4!*2!*4!), which is 34650.

a-i) There are 2*10! permutations starting with P, and each distinct word in the list will again be counted 4!*2!*4! times, so the total number of distinct permutations starting with P is 2*10!/(4!*2!*4!), which is 6300.

a-ii) In this case, we can consider the block SSSS to be a single symbol, in which case MISSISSIPPI contains eight symbols - one SSSS, 4 Is, 2Ps, and one M. Therefore, the total number of distinct permutations is 8!/(4!*2!), which is 840.

b) There is a bijection between the 10-letter words that can be formed and the 11-letter words, given by appending the one remaining letter to the end of the 10-letter word. Therefore, the total number of 10-letter words is simply 34650.

b-i) Considering the IIII as a single symbol, and using the reasoning in a-ii, there are 840 11-letter words formed having all four Is together. However, not all of them can be truncated to form valid 10-letter words, since if IIII is the last symbol, removing it creates a 7-letter word. We must therefore remove those 11-letter words containing the block IIII as the last symbol. There is one of these for each way of permuting the remaining 7 characters, of which there are 7!/(4!*2!) or 105. Therefore, the number of valid 10-letter words is 840-105 or 735.

Answer 2

We have a formula that says the number of ways n letters can be arranged is n!
If we want different ones and the letters repeat, then we divide by te factorial of the number of objects that repeat.

MISSISSIPPI has four I's, four S's, and two P's
( so to count for the repetition, we must divide by 4!4!2! )


a) 11! / [ 4! 2! 4! ]

i) 10! / [ 4! 4! ] (think of 10 choices and two groups of 4 repeat)


ii) 8! / [ 2! 4! ] Think of all SSSS as one object to be picked, but you have choices where to put it. We have the seven letter plus the SSSS makes 8!, divide by the repeated I's and the repeated PP

11! / [ 2! 4! 4! ] same as above we have 11 choose10 choices divided by repeats

11! / (11-8)![2! 4! ] again, think of IIII as one object, so the other 6 will give us 8 we have to choose from 11, and then divide by repeats.

Answer 3

i do no longer think that's named an anagram because of the fact an anagram makes use of the letters in any order many times mixing them up. for this reason the question is asked making use of the letters precisely of their order, i.e., group jointly anybody Achieves Miracles. i think of which would be distinctive than an anagram. i've got faith acrostic could be superb.