How many permutations of lowercase letters a bz contain at l

How many permutations of lowercase letters a, b,...,z contain at least one of the words wok, piwo, or okapi?. Write a complete justification of your answer.

Solution

there are 26 letters. these 26 letters can be arranged in 26! ways.

the number of permutations which contain at least one of the words wok,piwo,okapi=number of permutations containing wok+number of permutations containing piwo+number of permutations containing okapi-number of permutations containing both wok and piwo-number of permutations containing both wok and okapi-number of permutations containing both piwo and okapi+number of permutations containing all three

consider the word \"wok\" as a single entity. then there are rest 26-3=23 letters and this enity. so all total 24 elements which can be arranged in 24! ways.

consider the word \"piwo\" as a single entity. then there are rest 26-4=22 letters and this enity. so all total 23 elements which can be arranged in 23! ways.

consider the word \"okapi\" as a single entity. then there are rest 26-5=21 letters and this enity. so all total 22 elements which can be arranged in 22! ways.

now for both wok and piwo to exist we should have \"piwok\" as a single entity. then there are rest 26-5=21 letters and this enity. so all total 22 elements which can be arranged in 22! ways.

now for both wok and okapi to exist we should have \"wokapi\" as a single entity. then there are rest 26-6=20 letters and this enity. so all total 21 elements which can be arranged in 21! ways.

now for both piwo and okapi to exist we should have \"okapiwo\" as a single entity which is impossible because here \"o\" is repreated which can not be done. so both piwo and okapi can not be contained in any permuation. and hence all three can not be contained in any permuation at all

hence required answer=24!+22!+23!-22!-21!=24!+23!-21! [answer]       [where n!=n*(n-1)*(n-2)*....*3*2*1]