{"id":291,"date":"2024-04-01T14:02:19","date_gmt":"2024-04-01T13:02:19","guid":{"rendered":"https:\/\/tutors4you.com\/?page_id=291"},"modified":"2024-04-01T14:02:20","modified_gmt":"2024-04-01T13:02:20","slug":"restricted-permutations","status":"publish","type":"page","link":"https:\/\/tutors4you.com\/index.php\/restricted-permutations\/","title":{"rendered":"Restricted Permutations"},"content":{"rendered":"\n\n\t\t
\n\t\t\t\n\t\t\t

(a)   Number\n of permutations of ‘n’ things, taken\n ‘r’ at a time, when a particular thing is to be\n always included in each arrangement <\/font><\/p>\n

= r n-1<\/sup> Pr-1<\/sub><\/font><\/p>\n

(b) Number of\n permutations of ‘n’ things, taken ‘r’\n at a time, when a particular thing is fixed: = n-1<\/sup>\n Pr-1<\/sub><\/font><\/p>\n

(c) Number of\n permutations of ‘n’ things, taken ‘r’\n at a time, when a particular thing is never taken: = n-1<\/sup>\n Pr.<\/sub><\/font><\/p>\n

(d) Number of\n permutations of ‘n’ things, taken ‘r’\n at a time, when ‘m’ specified things always\n come together = m!  x (  n-m+1) !<\/font><\/p>\n

(e) Number of\n permutations of ‘n’ things, taken all at a\n time, when ‘m’ specified things always come\n together = n ! – [ m! x   (n-m+1)! ]<\/font><\/p>\n

Example:   How\n many words can be formed with the letters of the word\n ‘OMEGA’ when:<\/font><\/p>\n

(i)       ‘O’\n and ‘A’ occupying end places.<\/font><\/p>\n

(ii)       ‘E’\n being always in the middle<\/font><\/p>\n

(iii)       Vowels\n occupying odd-places<\/font><\/p>\n

(iv)        Vowels\n being never together.<\/font><\/p>\n

Ans. <\/font><\/p>\n

(i)    When\n ‘O’ and ‘A’ occupying end-places<\/font><\/p>\n

  => M.E.G.\n (OA)<\/font><\/p>\n

  Here (OA) are\n fixed, hence M, E, G can be arranged in  3! ways<\/font><\/p>\n

 But (O,A) can be\n arranged themselves is 2! ways.<\/font><\/p>\n

=> Total number of words\n =  3!   x  2! = 12 ways.<\/font><\/p>\n

(ii)  When\n ‘E’ is fixed in the middle<\/font><\/p>\n

=>   \n O.M.(E), G.A.<\/font><\/p>\n

Hence four-letter O.M.G.A.\n can be arranged in  4!   i.e 24 ways.<\/font><\/p>\n

(iii)   Three\n vowels (O,E,A,) can be arranged in the odd-places (1st<\/sup>,\n 3rd<\/sup> and 5th<\/sup>)   \n =  3!    ways.<\/font><\/p>\n

And two consonants (M,G,)\n can be arranged in the\n even-place              \n (2nd<\/sup>, 4th<\/sup>) =   2\n !   ways<\/font><\/p>\n

=> Total number of\n ways= 3! x 2! = 12 ways.<\/font><\/p>\n

(iv)  Total number\n of words   =   5!  \n =    120!<\/font><\/p>\n

 If all the vowels come\n together, then we have: (O.E.A.), M,G<\/font><\/p>\n

 These can be arranged\n in    3!    ways.<\/font><\/p>\n

 But (O,E.A.) can be\n arranged themselves in   3! ways.<\/font><\/p>\n

 => Number of ways,\n when vowels come-together  =   \n 3!  x    3!   <\/font><\/p>\n

= 36 ways<\/font><\/p>\n

=> Number of ways, when\n vowels being never-together <\/font><\/p>\n

=\n 120-36        = \n 84 ways.<\/font><\/p>\n

Number of Combination of\n ‘n’ different things, taken ‘r’ at a\n time is given by:-<\/font><\/p>\n

n<\/sup>Cr<\/sub>=  n!\n \/ r ! x\n (n-r)!                                \n <\/font><\/p>\n

Proof: Each combination\n consists of ‘r’ different things, which can be\n arranged among themselves in  \n r!  ways.<\/font><\/p>\n

=> For one combination of\n ‘r’ different things, number of arrangements\n =    r!<\/font><\/p>\n

For n<\/sup>Cr<\/sub>\n combination number of\n arrangements:     \n r    n<\/sup>Cr<\/sub><\/font><\/p>\n

=> Total number of\n permutations =    r!   n<\/sup>Cr<\/sub>\n  —————(1)  <\/font><\/p>\n

But number of permutation of\n ‘n’ different things, taken ‘r’ at a\n time<\/font><\/p>\n

= n<\/sup>Pr<\/sub>\n ——-(2)<\/font><\/p>\n

From (1) and (2) :<\/font><\/p>\n

n<\/sup>Pr<\/sub> \n =      r!  .  n<\/sup>Cr<\/sub><\/font><\/p>\n

or     \n n!\/(n-r)!  =  r!   .  n<\/sup>Cr   \n <\/sub>      <\/font><\/p>\n

or   n<\/sup>Cr<\/sub>   \n =       n!\/r!x(n-r)!<\/font><\/p>\n

Note: <\/strong>n<\/sup>Cr<\/sub> \n =  n<\/sup>Cn-r<\/sub><\/font><\/p>\n

or   n<\/sup>Cr<\/sub>   \n = n!\/r!x(n-r)!   and  n<\/sup>Cn-r<\/sub>  =   n!\/(n-r)!x(n-(n-r))!<\/font><\/p>\n

 =  n!\/(n-r)!xr!<\/font><\/p>\n

Restricted –\n Combinations<\/u><\/font><\/p>\n

(a)  Number of\n combinations of ‘n’ different things taken\n ‘r’ at a time, when ‘p’ particular\n things are always included = n-p<\/sup>Cr-p<\/sub>.<\/font><\/p>\n

(b)  Number of\n combination of ‘n’ different things, taken\n ‘r’ at a time, when ‘p’ particular\n things are always to be excluded = n-p<\/sup>Cr<\/sub><\/font><\/p>\n

Example:     \n In how many ways can a cricket-eleven be chosen out of 15\n players? if<\/font><\/p>\n

(i)  A particular\n player is always chosen,<\/font><\/p>\n

(ii)  A particular\n is never chosen.<\/font><\/p>\n

Ans:    <\/font><\/p>\n

(i)      \n A particular player is always chosen, it means that 10\n players are selected out of the remaining 14 players.<\/font><\/p>\n

=. Required number of ways\n =  14<\/sup>C10<\/sub>  = 14<\/sup>C4<\/sub><\/font><\/p>\n

= 14!\/4!x19!  =\n 1365                                              \n <\/font><\/p>\n

(ii) A\n particular players is never chosen, it means that 11\n players are selected out of 14 players.<\/font><\/p>\n

 => Required number\n of ways =  14<\/sup>C11<\/sub>  <\/font><\/p>\n

 =   14!\/11!x3!  =\n 364<\/font><\/p>\n

(iii) Number of ways of\n selecting zero or more things from ‘n’\n different things is given by:-   2n<\/sup>-1<\/font><\/p>\n

Proof<\/strong>:<\/font>  <\/font>Number of ways of selecting\n one thing, out of n-things     = n<\/sup>C1<\/sub><\/font><\/p>\n

Number of selecting two\n things, out of n-things =n<\/sup>C2<\/sub><\/font><\/p>\n

Number of ways of selecting\n three things, out of n-things =n<\/sup>C3<\/sub><\/font><\/p>\n

Number of ways of selecting\n ‘n’ things out of ‘n’ things = n<\/sup>Cn<\/sub><\/font><\/p>\n

=>Total number of ways of\n selecting one or more things out of n different things <\/font><\/p>\n

= n<\/sup>C1<\/sub>\n + n<\/sup>C2<\/sub> + n<\/sup>C3<\/sub> +\n ————- + n<\/sup>Cn<\/sub><\/font><\/p>\n

= (n<\/sup>C0<\/sub>\n + n<\/sup>C1<\/sub> + —————–n<\/sup>Cn<\/sub>) \n – n<\/sup>C0<\/sub><\/font><\/p>\n

= 2n<\/sup> –\n 1                          \n [ n<\/sup>C0<\/sub>=1]<\/font><\/p>\n

Example:  John has\n 8 friends. In how many ways can he invite one or more of\n them to dinner?<\/font><\/p>\n

Ans.    John\n can select one or more than one of his 8 friends.<\/font><\/p>\n

=> Required number of\n ways = 28<\/sup> – 1= 255.<\/font><\/p>\n

(iv) Number of ways of\n selecting zero or more things from ‘n’\n identical things is given by :- n+1<\/font><\/p>\n

Example:   In how\n many ways, can zero or more letters be selected form the\n letters AAAAA?<\/font><\/p>\n

Ans. Number of ways of :<\/font><\/p>\n

        \n Selecting zero ‘A’s   =  1<\/font><\/p>\n

        \n Selecting one ‘A’s   =   1<\/font><\/p>\n

        \n Selecting two ‘A’s     = 1<\/font><\/p>\n

        \n Selecting three ‘A’s  =  1<\/font><\/p>\n

        \n Selecting four ‘A’s   =   1<\/font><\/p>\n

        \n Selecting five ‘A’s    =  1<\/font><\/p>\n

=> Required number of\n ways  =\n 6         [5+1]<\/font><\/p>\n

(V)    Number\n of ways of selecting one or more things from\n ‘p’ identical things of one type ‘q’\n identical things of another type, ‘r’ identical\n things of the third type and ‘n’ different\n things is given by :-<\/font><\/p>\n

 (p+1) (q+1) (r+1)2n<\/sup>\n – 1<\/font><\/p>\n

Example:    Find\n the number of different choices that can be made from 3\n apples, 4 bananas and 5 mangoes, if at least one fruit is\n to be chosen.<\/font><\/p>\n

Ans: <\/font><\/p>\n

Number of ways of selecting\n apples = (3+1) = 4 ways.<\/font><\/p>\n

Number of ways of selecting\n bananas = (4+1) = 5 ways.<\/font><\/p>\n

Number of ways of selecting\n mangoes = (5+1) = 6 ways.<\/font><\/p>\n

Total number of ways of\n selecting fruits = 4 x 5 x 6<\/font><\/p>\n

But this includes, when no\n fruits i.e. zero fruits is selected<\/font><\/p>\n

=> Number of ways of\n selecting at least one fruit = (4x5x6) -1  = 119<\/font><\/p>\n

Note :- There was no fruit\n of a different type, hence here  n=o<\/font><\/p>\n

 =>   2n \n <\/sup>= 20<\/sup>=1<\/font><\/p>\n

(VI)   Number of\n ways of selecting ‘r’ things from ‘n’\n identical things is ‘1’.<\/font><\/p>\n

Example:   In how\n many ways 5 balls can be selected from ‘12’\n identical red balls?<\/font><\/p>\n

Ans. The balls are\n identical, total number of ways of selecting 5\n balls  = 1.<\/font><\/p>\n

Example: How many numbers of\n four digits can be formed with digits 1, 2, 3, 4 and 5?<\/font><\/p>\n

Ans. Here n =\n 5                    \n [Number of digits]<\/font><\/p>\n

And   r =\n 4                    \n [ Number of places to be filled-up]<\/font><\/p>\n

Required number is   5<\/sup>P4<\/sub>\n = 5!\/1!  = 5 x 4 x 3 x 2 x 1<\/font><\/p>\n

                                                           \n <\/font><\/p>\n \n\t\t<\/div>\n\t\n","protected":false},"excerpt":{"rendered":"

(a)   Number of permutations of ‘n’ things, taken ‘r’ at a time, when a particular thing is to be always included in each arrangement = r n-1 Pr-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’ […]<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":0,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"","meta":{"footnotes":""},"class_list":["post-291","page","type-page","status-publish","hentry"],"blocksy_meta":[],"yoast_head":"\nRestricted Permutations - Tutors 4 You<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/tutors4you.com\/index.php\/restricted-permutations\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Restricted Permutations - 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