Question

In word BOOKKEEPER, how many distinct arrangements are there if the letter P must always occur before any vowels?

Answer 1

BOOK KEEPER We want to find no. of distinct arrangeme -nt such that p must always orcur before any vowel. We use mutinomial number :- let s be set of n object n, 102,.., am no such that hit n2+...+ ne zn. mutinomial no (n is definde to be h na ... hm) distribution of ith type of object with Di numbers (idiom) in boxes. formula :-/ ini me... hm ) ha! ha!... ni

In BOOKKEEPER there are 5 vowels We want to find urrangement in such a way that P oceurs before any vowel. ie. P comes together with any vowel. 1) P comes together with which is at second plere le. B POOKEE ER Here we treat Po as one letter. So no. of arrangement by multinomial no.) : 8! 1!1!1!1!3!1! = 8X 7 X6 x 5 xy = 6720 1) P comes together with 'o' which is at 3rd place No. of or rangengent = 6720

e comes together with Eie. SPE OOKKEEL No. of arrangement = 3! -= 5040 2. 2! 2! There are 3 times E. So no. of arrangen for all E= 3x 5040 = 20 120 Total no. of arrangent (fron () 92(1) = 6720 + 6720 + 20120 = 33 560