BY THE SYMBOL n! ("n factorial") we mean the product of consecutive numbers 1 through n.
=n x (n - 1) x (n - 2) x . . . .x 3 x 2 x 1 The order of the factors does not matter, whether backwards or forwards.
Counting Principles Fundamental Counting Principle 1: Fundamental Counting Principle 2:
BY THE PERMUTATIONS of the letters abc we mean all of their possible arrangements: abc acb bac bca cab cba There are 6 permutations of three different things The arrangements of n elements taken m at a time represent, in fact, a partial permutation. Older books use the notation nPm (n Permutation m) for arrangements: The number of permutations of n different things taken n at a time is n!. a) Number of permutations of n things, taken r at a time, when a particular thing is to be always included in each arrangement = rn-1Pr-1 (b) Number of permutations of n things, taken r at a time, when a particular thing is fixed: = n-1 Pr-1 (c) Number of permutations of n things, taken r at a time, when a particular thing is never taken: = n-1 Pr. (d) Number of permutations of n things, taken r at a time, when m specified things always come together = m! x ( n-m+1) ! (e) Number of permutations of n things, taken all at a time, when m specified things always come together = n ! - [ m! x (n-m+1)! ] (f) The number of permutations of n elements with n1 of repeated element, n2 of another repeated element, etc. taken n at a time. (g) The number of arrangement of n different things, taken r at a time, when each may be repeated any number of times in each arrangement, is nr
There are two cases of circular-permutations: -
(b) If clock-wise and anti-clock-wise orders are taken as not different, then total number of circular-permutations is given by (n-1)!/2! Number of circular-permutations of n different things taken r at a time:- (a) If clock-wise and anti-clockwise orders are taken as different, then total number of circular-permutations = nPr /r (b) If clock-wise and anti-clockwise orders are taken as not different, then total number of circular permutation = nPr/2r Examples Q1. Your state issues license plates consisting of letters and numbers. There are 26 letters and the letters may be repeated. There are 10 digits and the digits may be repeated. How many possible license plates can be issued with two letters followed by three numbers? Q2. How many different signals can be made by 5 flags from 8-flags of different colours? Q3. Find the total number of words that can be made by using all the letters from the word ‘MACHINE’ using only once. Q4. In how many ways can the letters of the word ABACUS be rearranged such that the vowels always appear together? Q5. In how many different ways, can 4 boys and 4 girls can arrange themselves in a circle such that boys & girls alternate? Q6. How many words can be made by using the letters of the word "SIMPLETON" taken all at a time? Q7. How many different 4-dgit numbers can be made from the first 4 natural numbers, using each digit only once? Q8. Heather has finally narrowed her clothing choices for the big party down to 3 skirts, 2 tops and 4 pair of shoes. How many different outfits could she form from these choices? Q9. In how many ways can 5 letters be posted in 3 post boxes, if any number of letters can be posted in all of the three post boxes? |
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I have got certain doubts in the answers to questions 4 & 5.Here were my solutions -
Q4) Since AAU are the vowels in ABACUS & they form a group, we have 4 elements to arrange ( AAU,B,C,S). These can be arranged in 4! ways.
Also the elements in AAU, themselves can be arranged in 3! ways.
So the answer is 4! x 3! ways. How does the 2 in the denominator come in picture, since i believe there aren't any common terms.
Q5)The girls & boys can be arranged as follows
G
B
G
B
G
B
G
B
.
Thus, the girls can be arranged in 4 places in 4! ways & so can the boys. But,since over here we have considered the starting place to be that of a girl, which can be occupied by a boy, we should multiply the resultant by 2.
The ans thus, is 4! x 4! x 2.
Can anyone help me figure out where i am going wrong(assuming the solutions provided are correct).
About the Q4, with AAU you have 3!/2! permutations, so, in the end you have 4! permutations with the "letters" AAU,B,C,U,S but you have to account for only the ones in AAU that are not repeated (ie, A1,A2,U == A2,A1,U).
4!x(3!/2!) = (4!x3!)/2! = (4!x3!)/2
for the fourt question we divde it by two because there are two similar vowels a