{"id":3151,"date":"2023-10-04T12:09:15","date_gmt":"2023-10-04T12:09:15","guid":{"rendered":"http:\/\/localhost\/ecole9ja\/?p=3151"},"modified":"2023-10-04T12:23:00","modified_gmt":"2023-10-04T12:23:00","slug":"week-9-ss2-second-term-further-mathematics-notes","status":"publish","type":"post","link":"https:\/\/ecolebooks.com\/nigeria\/posts\/week-9-ss2-second-term-further-mathematics-notes\/","title":{"rendered":"Week 9 – SS2 Second Term Further Mathematics Notes"},"content":{"rendered":"

\u00a0WEEK NINE<\/strong>
\n\t\tTOPIC:PERMUTATION AND COMBINATION :
\n<\/strong>PERMUTATION:Definitoin, Concept , Different Arrangement Of Items, Cyclic Permutation .
\n<\/strong><\/p>\n

    \n
  1. \n
    Definition, Concept:\n<\/div>\n

    Definition: Permutation is defined as the number of arranged of objects. The different orders of arrangement are important. E.g. Find the number of ways of arranging the letters pqr.
    \nPqr, prq, qrp, qpr, rqp. The number of ways is 6 ways
    \nSimilarly, for 4 letters the number of arrangement is 24
    \nIn general, the number of different arrangement of n different objects is equal to n! (n factorial)
    \nN! = n x (n-1) x (n-2) x \u2026 x 3×2 x 1×0!\u00a0\u00a0\u00a0\u00a0(But, 0! = 1)\n<\/li>\n

  2. \n
    Simplify the following: A. 5!\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0B.\n<\/div>\n

    Solution
    \n<\/strong><\/li>\n

  3. \n
    5! = 5x4x3x2x1 = 120\n<\/div>\n<\/li>\n
  4. \n
    = 7x6x5x4! = 7×5 = 35\n<\/div>\n<\/li>\n
  5. \n
    Find the number of ways of arranging the letters of the word MACHINE\n<\/div>\n

    Solution:
    \n<\/strong>There are seven different letters in the word MACHINE, therefore the number of permutation is 7! = 7x6x5x4x3x2x1 = 5040 ways\n<\/li>\n

  6. \n
    Simplify (n + 1)! = (n+1)n! = n+1\n<\/div>\n

    (n-1)!\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0(n-1)n!\u00a0\u00a0\u00a0\u00a0n-1<\/p>\n

    \u00a0ARRANGEMENT OF n-OBJECTS TAKING r-OBJECTS
    \nIf we are interested in the number of ways 2 letters of a 4 lettered word cnbe arranged, then the npr<\/sub> is the permutation of n objects taking at a time
    \nn<\/sup>pr<\/sub> =
    \nExample: Evaluate: \u00a0\u00a0\u00a0\u00a0(a) 8p3<\/sub>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0(b) 11p9
    \n<\/sub>Solution:
    \n<\/sub><\/li>\n

  7. \n
    8<\/sup>P3<\/sub> = = = = 8x7x6 = 336\n<\/div>\n<\/li>\n
  8. \n
    11<\/sup>P9<\/sub>= = 11x0x9x8x7x6x5x4x3x2! = 19958400\n<\/div>\n<\/li>\n
  9. \n
    In how many ways can three people be seated on eight seats in a row?\n<\/div>\n

    Solution:
    \n1st<\/sup> seat can be occupied by any of the 8 = 8 ways
    \n2nd<\/sup> seat can be occupied in 6 ways
    \nHence, the number of ways = 8x7x6 = 336 ways
    \nAlternatively, n = 8, r = 3
    \nn<\/sup>Pr<\/sub> = = = = 8x7x6 = 336 ways<\/p>\n

    \u00a0EVALUATION
    \n<\/strong><\/li>\n

  10. \n
    In how may ways can 8 students be seated in a row?\n<\/div>\n<\/li>\n
  11. \n
    In how many ways can the 1st<\/sup>, 2nd<\/sup> 3rd<\/sup> prizes be won by 6 athletetes in a race?\n<\/div>\n<\/li>\n
  12. \n
    In how many ways can the letters of the word HISTORY be arranged?\n<\/div>\n

    \u00a0CYCLIC PERMUTATION:<\/strong> Cyclic permutation is the arrangement of things around a circular object. Since a circular table has no beginning and no end, the number of arrangement is 1 x (n \u2013 1) !
    \nIf the circular object can be turned over e.g. circular ring e.t.c. the number of arrangement =
    \nExample: In how many ways can 6 members of a disciplinary committee be seated round a circular table?<\/p>\n

    \u00a0Solution:
    \n<\/strong>The number of ways = (n \u2013 1)! X 1
    \nN = 6,
    \nHence, (6 \u2013 1)! X 1 = 5! X 1 = 120 ways
    \nPERMUTATION OF IDENTICAL OBJECTS:
    \nThe number of ways of permuting n objects taking n at a time with n, objects alike, n2<\/sub>alke is, <\/p>\n<\/li>\n

  13. \n
    Find the number of ways the word MATHEMATICS can be arranged.\n<\/div>\n

    Solution:<\/p>\n

    \u00a0MATHEMATICS
    \n<\/strong>There are: 2Ms, 2Asm 2Ts and 11 letters.
    \nN=11, n1 = 2! N2 = 2! N3 = 2
    \n = 11x10x9x7x6x5x4x3x2x1 = 4989600<\/p>\n

    \u00a0CONDITIONAL PERMUTATION:
    \n<\/strong>Sometimes restrictions are placed on the order of arrangements of objects
    \nExamples:
    \n<\/strong><\/li>\n

  14. \n
    Find the number of ways the letters of the word COMMITTEE can be permuted, if the 2Ts must always be together.\n<\/div>\n

    Solution:
    \n<\/strong>The 2Ts must be together, we can lump them as follows: COMMI (TT) EE = 8!
    \n= = 10080 ways\n<\/li>\n

  15. \n
    Find the number of ways of arranging the letters of the word MOSHOESHOE if the letter M must always begin a word\n<\/div>\n

    Solution:
    \n<\/strong>Since letter m must always begin, and then m can only occupy the first position
    \ni.e M = 1 way
    \nother letters, OSHOESHOE = = 9x7x6x5x4 = 7560 ways<\/p>\n

    \u00a0COMBINATION: Selection, Conditional Selection And Its Application
    \n<\/strong>Combination can be defined as the number of ways r \u2013 objects can be selected from n \u2013 objects irrespective of the arrangement
    \nHence, the notation is thus, n<\/sup>Cr<\/sub> or (n<\/sup>r<\/sub>)
    \n, n<\/sup>Cr<\/sub>=
    \nRelationship between permutation and combination is thus, nCr =
    \nExample:
    \n<\/strong><\/li>\n

  16. \n
    Evaluate 10<\/sup>C4<\/sub>\n\t\t\t\t<\/div>\n

    Solution:
    \n<\/strong>10<\/sup>C4 =<\/sub> = 10 x 3 x 7 = 210
    \n\t\t\t\t\t<\/sup><\/li>\n

  17. \n
    In how many ways can three books be selected from 12 books?\n<\/div>\n

    SOLUTION:
    \n<\/strong>N = 12, r = 3, 12<\/sup>C3<\/sub> = = 2x11x10 = 220 ways<\/p>\n

      \n
    1. \n
      A committee consisting of 3 men and 5 women is selected from 5 men and 10 women. Find how many ways this committee can be formed.\n<\/div>\n<\/li>\n<\/ol>\n

      Solution:
      \n<\/strong>MEN\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0WOMEN
      \n<\/strong>R = 3, n = 5\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0r = 5, n = 10
      \n5<\/sup>C3<\/sub> = \u00a0\u00a0\u00a0\u00a0 = 10\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a010<\/sup>C5<\/sub> = = 252
      \nTherefore number of ways of selecting the committee = 10×252 = 2520 ways.<\/p>\n

      \u00a0GENERAL\/ REVISION EVALUATION
      \n<\/strong><\/li>\n

    2. \n
      Find the number of ways the letters of the word FURTHER can be arranged\n<\/div>\n<\/li>\n
    3. \n
      Find the number of ways of arranging 7 people in a straight line, if two particular people must always be separated\n<\/div>\n<\/li>\n
    4. \n
      In how many ways can 6 pupils be lined up if 3 of them insist in the following one another\n<\/div>\n<\/li>\n
    5. \n
      Verify that = (n \u2013 1) (n \u2013 2) (n \u2013 3)!\n<\/div>\n

      \u00a0READING ASSIGNMENT
      \n<\/strong>Read permutation and combination, further mathematics project 2 pages 47-54<\/p>\n

      \u00a0WEEKEND ASSIGNMENT
      \n<\/strong><\/p>\n

        \n
      1. \n
        Evaluate 6<\/sup>C2<\/sub> + 6<\/sup>C3<\/sub> + 6<\/sup>C4<\/sub> + 6<\/sup>C5<\/sub> (a) 6<\/sup>C6<\/sub> (b) 6<\/sup>C5<\/sub> (c) 8<\/sup>C5<\/sub>\n\t\t\t\t\t\t<\/div>\n<\/li>\n
      2. \n
        How much ways can the letters of the word EVALUATE be arranged? (a) 10080 (b) 20160 (c) 40320\n<\/div>\n<\/li>\n
      3. \n
        In how many ways can 2 boys and 3 girls be arranged to sit in a row, if the boys must sit together (a) 6 (b) 4 (c) 24\n<\/div>\n<\/li>\n
      4. \n
        Find the number of ways 6 people can be seated in a round table, if two particular friends must sit next to each other (a) 48 9b) 24 (c) 120\n<\/div>\n<\/li>\n
      5. \n
        In how many ways can 6 pupils be lined up if 3 of them insist on following one another? (a) 720 (b) 144 (c) 24\n<\/div>\n<\/li>\n<\/ol>\n

        \u00a0THEORY
        \n<\/strong><\/p>\n

          \n
        1. \n
          Out of 7 lawyers, 5 judges, a committee consisting of 3 lawyers, 2 judges is to be formed, in how many ways can this be done, if\n<\/div>\n
            \n
          1. \n
            Any lawyer and any judge can be included\n<\/div>\n<\/li>\n
          2. \n
            One particular judge can be included\n<\/div>\n<\/li>\n
          3. \n
            Two particular lawyer cannot be in committee\n<\/div>\n<\/li>\n<\/ol>\n<\/li>\n<\/ol>\n<\/li>\n
          4. \n
            If n<\/sup>P3<\/sub> \/ n<\/sup>C2<\/sub> = 6, find the value of n\n<\/div>\n


            \n\t\t\t\t\t<\/strong>\u00a0<\/li>\n<\/ol>\n","protected":false},"excerpt":{"rendered":"

            \u00a0WEEK NINE TOPIC:PERMUTATION AND COMBINATION : PERMUTATION:Definitoin, Concept , Different Arrangement Of Items, Cyclic Permutation…<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1,253],"tags":[],"class_list":["post-3151","post","type-post","status-publish","format-standard","hentry","category-posts","category-second-term-ss2-further-mathematics"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts\/3151","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/comments?post=3151"}],"version-history":[{"count":1,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts\/3151\/revisions"}],"predecessor-version":[{"id":3152,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts\/3151\/revisions\/3152"}],"wp:attachment":[{"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/media?parent=3151"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/categories?post=3151"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/tags?post=3151"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}