Expert: Scotto - 1/27/2008

Question
hi i've been trying these Permutation questions and would like to know if I am going right so far.Also where I have used incorrect notation could you please show me where I have gone wrong. Qu1 A box contains 8 cards,of which 5 are labelled A and 3 are labelled B.Find (i) the number of different permutations of the 8 cards(ii)the number of these permutations beginning with three A's; ANS(i)8 cards chosen 2 ways=8P2=8*7=56;(ii)with 3 A's at beginning there are 5 cards left which can either still be A or B->5P2=5*4=20; Qu2.(i) Find the total number of different permutations of all the letters of the word RESERVE(ii)Find the number of permutations in which E is the 1st letter(iii)find the number of permutations in which the two R's come together(iv)Find the number of permutations in which S and V come at the ends ANS:(i)Permutations of the word RESERVE=7!=5040(ii)E is the 1st letter therefore 6 spaces are left->6!=720(iii)2 R's come together there 5 spaces left so we have 5!=120 (iv)S and V come at the ends so we have 6!*2=720 since SV at the ends can also be written VS.Qu3 A row of ten houses are to be painted in 3 colours. 2 houses are to be red,3 to be blue and 5 to be white. Find the no. of diff. ways in which the houses can be painted (i)with no restrictions(ii)given that the 1st and last houses in the row are blue(iii)given that the 1st and last houses in the row are the same colour  ANS (i)10!=3628800 (ii)9!=362880(iii)9!/6!=504

Answer
You look like this can help you.

On the first question, given that a box contains 8 cards with n A and m B, the number of permutations is 8!/(n!m!).

If one card is put in a special place, say an A, then there are only 7 cards left with (n-1) A cards.  The answer for this is then (8-1)!/((n-1)!m!.  If two cards of a particular color, say A, are to be together, then treat the problem as if there were only 8-1=7 cards with (n-1) A cards.

On the second question, with a subset of 4 items in a subset of n, we need to know how many of each item there are, say n1, n2, n3, and n4.  The total items is S=n1+n2+n3+n4.  If we want to know the number of ways to organize them, its S!/(n1!n2!n3!n4!).

Grouping two items together, say from group 2, forces us to reconsider the problem as if there were S-1 items and n2-1 items from group 2.  The number of arrangements is then
(S-1)!/(n1!(n2-1)!n3!n4!)

When the set is split into groups 1 and 2 being at the start and groups 3 and 4 being at the end, we need to find S1=n1+n2 and S2=n3+n3.  The number of permutations is then S1!/(n1!n2!) + S2!/(n3!n4!).

Using this foundation, the number of permutations of 8 cards is 8!.  The number of permutations starting with 3 A’s is (3!/3!)(5!/2!3!).  The word RESERVE has 2-R’s,
4-E’s, 1 S, and 1 V.  If the letters are drawn at random, the chance of getting them in the right order is 2!3!1!1!/8!.  If 2-Es are grouped together, the problem should be reconsidered as if there were only 3-E’s and 7 letters.  If the vowels are considered to be at the start the number of ways is (3!/3!)(4!/(2!1!1!)) = (6/6)(24/2) = 12.

Apply this concept to the questions at hand.