Permutations Question

Question

1. a) How many permutations are there of the letters of the word BASKETBALL? I know the answer is 453,600. (Needed help for b,c, and d)

b) How may of the arrangements begin with K?

c) How many of the arrangements start with B?

d) In how many of the arrangements would the two L's be together?

Explain please in detail.

Answer 1

Yes, you are right on the first part.

(b) Since K is fixed, you don't have to include K in your arrangement. There are 9 letters you need to arrange including 2 B's, 2 A's and 2 L's

(9!)/(2!2!2!) = 45360

(c) Since one B is fixed, you don't have to include one B in your arrangement. That means there are 9 letters you need to arrange including 2 A's and 2 L's

(9!)/(2!2!) = 90720

(d)This case is a little different. You can have 9 formats like this:

LL_ _ _ _ _ _ _ _

_LL_ _ _ _ _ _ _

_ _LL_ _ _ _ _ _

_ _ _LL _ _ _ _ _

_ _ _ _LL _ _ _ _

_ _ _ _ _LL _ _ _

_ _ _ _ _ _LL _ _

_ _ _ _ _ _ _LL _

_ _ _ _ _ _ _ _LL

What you need to do is to get the permutation of one format (like we did in b and c) then, we multiply by 9.

8!/(2!2!) * 9 = 90720

Answer 2

All of these parts of the question are variations using the same formula: (total # of letters)! / (# of repeated letters)!

So a) should be 10!/(2!2!2!), which I assume is the 453,600 you cite.

b) If K is the 1st letter, then we are arranging the remaining 9 letters, with the same repeats: 9!/(2!2!2!)

c) With B 1st, we arrange the remaining 9 but now the other B is not a repeat: 9!/(2!2!)

d) The pair of consecutive L's can be in any of 9 spots. The remaining 8 letters, with 2 repeats, can be permuted in the usual fashion: 9·8!/(2!2!)

Answer 3

(b) If the first letter is K, the other 9 letters can be arranged in 9! / [2! 2! 2!] ways.

(c) If the first letter is B, the other 9 can be arranged in 9! / [2! 2!] ways

(d) We can place the LL as follows

LLxxxxxxxx

xLLxxxxxxx

xxLLxxxxxx

xxxLLxxxxx

xxxxLLxxxx

xxxxxLLxxx

xxxxxxLLxx

xxxxxxxLLx

xxxxxxxxLL

(x is one of the other 8 letters)

In each of these 9 cases, the other 8 letters can be placed in 8 ! / [2! 2!] ways

So total arrangements = 9 x [8! / [2! 2!]]