{"id":19867,"date":"2022-08-09T17:29:53","date_gmt":"2022-08-09T11:59:53","guid":{"rendered":"https:\/\/apboardsolutions.com\/?p=19867"},"modified":"2022-08-11T14:58:16","modified_gmt":"2022-08-11T09:28:16","slug":"inter-2nd-year-maths-2a-permutations-and-combinations-formulas","status":"publish","type":"post","link":"https:\/\/apboardsolutions.com\/inter-2nd-year-maths-2a-permutations-and-combinations-formulas\/","title":{"rendered":"Inter 2nd Year Maths 2A Permutations and Combinations Formulas"},"content":{"rendered":"<p>Use these <a href=\"https:\/\/apboardsolutions.com\/inter-2nd-year-maths-2a-formulas\/\">Inter 2nd Year Maths 2A Formulas<\/a> PDF Chapter 5 Permutations and Combinations to solve questions creatively.<\/p>\n<h2>Intermediate 2nd Year Maths 2A Permutations and Combinations Formulas<\/h2>\n<p>\u2192 Fundamental principle: If a work can be performed in m different ways and another work can be performed in n different ways, then the two works simultaneously can be performed in mn different ways.<\/p>\n<p>\u2192 If n is a non-negative integer then<\/p>\n<ul>\n<li>0 ! = 1<\/li>\n<li>n ! = n (n &#8211; 1) ! if n &gt; 0<\/li>\n<\/ul>\n<p>\u2192 The number of permutations of n dissimilar things taken &#8216;r at a time is denoted by <sup>n<\/sup>P<sub>r<\/sub> and <sup>n<\/sup>P<sub>r<\/sub> = \\(\\frac{n !}{(n-r) !}\\) for 0 \u2264 r \u2264 n.<\/p>\n<p><img loading=\"lazy\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/07\/AP-Board-Solutions.png\" alt=\"Inter 2nd Year Maths 2A Permutations and Combinations Formulas\" width=\"196\" height=\"17\" \/><\/p>\n<p>\u2192 If n, r positive integers and r \u2264 n, then<\/p>\n<ul>\n<li><sup>n<\/sup>P<sub>r<\/sub> = n. <sup>n &#8211; 1<\/sup>P<sub>r &#8211; 1<\/sub> if r \u2265 1<\/li>\n<li><sup>n<\/sup>P<sub>r<\/sub> = n.(n &#8211; 1) <sup>n &#8211; 2<\/sup>P<sub>r &#8211; 2<\/sub> if r \u2265 2<\/li>\n<li><sup>n<\/sup>P<sub>r<\/sub> = <sup>n &#8211; 1<\/sup>P<sub>r<\/sub> + r. <sup>(n &#8211; 1)<\/sup>P<sub>(r &#8211; 1)<\/sub><\/li>\n<\/ul>\n<p>\u2192 The number of injections that can be defined from set A into set B is <sup>n(B)<\/sup>p<sub>n(A)<\/sub> n(A) \u2264 n(B)<\/p>\n<p>\u2192 The number of bijections that can be defined from set A into set B having same number of elements with A is n(A)!<\/p>\n<p>\u2192 The number of permutations of1n&#8217; dissimilar things taken &#8216;r&#8217; at a time.<\/p>\n<ul>\n<li>Containing a particular thing is (r) <sup>n &#8211; 1<\/sup>P<sub>r &#8211; 1<\/sub><\/li>\n<li>Not containing a particular thing is <sup>n &#8211; 1<\/sup>P<sub>r<\/sub><\/li>\n<li>Containing a particular thing in a particular place is <sup>n &#8211; 1<\/sup>P<sub>r &#8211; i<\/sub>.<\/li>\n<\/ul>\n<p>\u2192 The number of functions that can be defined from set A into set B in [n(B)]<sup>n(A)<\/sup><\/p>\n<p>\u2192 The sum of the r &#8211; digited numbers that can be formed using the given &#8216;n&#8217; distinct non-zero digits (r \u2264 n \u2264 9) is <sup>(n &#8211; 1)<\/sup>P<sub>(r &#8211; 1)<\/sub> \u00d7 sum of all n digits \u00d7 111 &#8230;&#8230; 1 (r times)<\/p>\n<p>\u2192 In the above, if &#8216;0&#8217; is one among the given n digits, then the sum is <sup>(n &#8211; 1)<\/sup>P<sub>(r &#8211; 1)<\/sub> \u00d7 sum of the digits \u00d7 111 &#8230; 1 (r times) <sup>(n &#8211; 2)<\/sup>P<sub>(r &#8211; 2)<\/sub> \u00d7 sum of the digits \u00d7 111 &#8230; 1 [(r &#8211; 1) times]<\/p>\n<p>\u2192 The number of permutations of n dissimilar things taken r at a time when repetitions are allowed any number of times is n<sup>r<\/sup>.<\/p>\n<p>\u2192 The number of circular permutations of n dissimilar things is (n &#8211; 1) !<\/p>\n<p>\u2192 In case of hanging type circular permutations like garlands of flowers, chains of beads etc., the number of circular permutations of n things is \\(\\frac{1}{2}\\) [(n &#8211; 1) !]<\/p>\n<p>\u2192 If in the given n things, p alike things are of one kind, q alike things are of the second kind, r alike things are of the third kind and the rest are dissimilar, then the number of permutations (of these n things) is \\(\\frac{n !}{p ! q ! r !}\\)<\/p>\n<p><img loading=\"lazy\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/07\/AP-Board-Solutions.png\" alt=\"Inter 2nd Year Maths 2A Permutations and Combinations Formulas\" width=\"196\" height=\"17\" \/><\/p>\n<p>\u2192 The number of combinations of n things taken r at a time is denoted by <sup>n<\/sup>C<sub>r<\/sub> and <sup>n<\/sup>C<sub>r<\/sub> = \\(\\frac{n !}{(n-r) ! r !}\\) for 0 \u2264 r \u2264 n and \\(\\frac{{ }^{n} C_{r}}{{ }^{n} C_{r-1}}=\\frac{n-r+1}{r}\\) and <sup>n<\/sup>P<sub>r<\/sub> = r! <sup>n<\/sup>C<sub>r<\/sub><\/p>\n<p>\u2192 If n, r are integers and 0 \u2264 r \u2264 n then <sup>n<\/sup>c<sub>r<\/sub> = <sup>n<\/sup>C<sub>r &#8211; 1<\/sub><\/p>\n<p>\u2192 Let n, r, s are integers and 0 \u2264 r \u2264 n, 0 \u2264 s \u2264 n. If <sup>n<\/sup>C<sub>r<\/sub> = <sup>n<\/sup>C<sub>s<\/sub> then r = s or r + s = n.<\/p>\n<p>\u2192 The number of ways of dividing \u2018m + n&#8217; things (m \u2260 n) into two groups containing m, n things is <sup>(m + n)<\/sup>C<sub>m<\/sub> = <sup>(m + n)<\/sup>C<sub>m<\/sub> = \\(\\frac{(m+n) !}{m ! n !}\\)<\/p>\n<p>\u2192 The number of ways of dividing (m + n + p) things (m, n, p are distinct) into 3 groups of m, n, p things is \\(\\frac{(m+n+p) !}{m ! n ! p !}\\).<\/p>\n<p>\u2192 The number of ways of dividing mn things into m equal groups is \\(\\frac{(m n) !}{(n !)^{m} m !}\\)<\/p>\n<p>\u2192 The number of ways if distributing mn things equally to m persons is \\(\\frac{(m n) !}{(n !)^{m}}\\)<\/p>\n<p>\u2192 If p alike things are of one kind, q alike things are of the second kind and r alike things are of the third kind, then the number of ways of selecting one or more things out of them is (p + 1) (q + 1) (r +1) &#8211; 1.<\/p>\n<p>\u2192 If m is a positive integer and m = p<sub>1<\/sub><sup>\u03b1<sub>1<\/sub><\/sup>, p<sub>1<\/sub><sup>\u03b1<sub>2<\/sub><\/sup> &#8230;&#8230; p<sub>k<\/sub><sup>\u03b1<sub>k<\/sub><\/sup> where p<sub>1<\/sub>, p<sub>2<\/sub> &#8230;&#8230; p<sub>k<\/sub> are distinct primes and \u03b1<sub>1<\/sub>, \u03b1<sub>2<\/sub>, &#8230;. \u03b1<sub>k<\/sub> are non-negative integers, then the number of divisors of m is (\u03b1<sub>1<\/sub> + 1) (\u03b1<sub>2<\/sub> + 1) &#8230;&#8230;&#8230; (\u03b1<sub>k<\/sub> + 1). [This includes 1 and m].<\/p>\n<p><img loading=\"lazy\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/07\/AP-Board-Solutions.png\" alt=\"Inter 2nd Year Maths 2A Permutations and Combinations Formulas\" width=\"196\" height=\"17\" \/><\/p>\n<p>\u2192 The total number of combinations of n different things taken any number of times is 2<sup>n<\/sup>.<\/p>\n<p>\u2192 The total number of combinations of n different things taken one or more at a timers 2<sup>n<\/sup> &#8211; 1.<\/p>\n<p>\u2192 The number of diagonals in a regular polygon of n sides is \\(\\frac{n(n-3)}{2}\\)<\/p>\n<p>\u2192 Permutations are arrangements of things taken some or all at a time.<\/p>\n<p>\u2192 In a permutation, order of the things is taken into consideration.<\/p>\n<p>\u2192 <sup>n<\/sup>p<sub>r<\/sub> represents the number of permutations (without repetitions) of n dissimilar things taken r at a time.<\/p>\n<p>\u2192 Fundamental Principle: If an event is done in \u2018m\u2019 ways and another event is done in \u2018n\u2019 ways, then the two events can be together done in mn ways provided the events are independent.<\/p>\n<p>\u2192 <sup>n<\/sup>p<sub>r<\/sub> = \\(\\frac{n !}{(n-r) !}\\) = n(n &#8211; 1)(n &#8211; 2) &#8230;&#8230;&#8230;&#8230;. (n &#8211; r + 1)<\/p>\n<p>\u2192 <sup>n<\/sup>p<sub>1<\/sub> = n, <sup>n<\/sup>p<sub>2<\/sub> = n(n-1), <sup>n<\/sup>p<sub>3<\/sub> = n(n-1)(n-2).<\/p>\n<p>\u2192 <sup>n<\/sup>p<sub>r<\/sub> = n!<\/p>\n<p>\u2192 \\(\\frac{{ }^{\\mathrm{n}} \\mathrm{p}_{\\mathrm{r}}}{{ }^{\\mathrm{n}} \\mathrm{p}_{\\mathrm{r}-1}}\\)<sup>n<\/sup>p<sub>n<\/sub> = n!<\/p>\n<p>\u2192 <sup>n+1<\/sup>p<sub>r<\/sub> = r. <sup>n<\/sup>p<sub>r-1<\/sub> + <sup>n<\/sup>p<sub>r<\/sub><\/p>\n<p>\u2192 <sup>n<\/sup>p<sub>r<\/sub> = r. <sup>n-1<\/sup>p<sub>r-1<\/sub> + <sup>n<\/sup>p<sub>r<\/sub><\/p>\n<p>\u2192 <sup>n<\/sup>p<sub>r<\/sub> = r.<sup>n-1<\/sup>p<sub>r-1<\/sub> + <sup>n-1<\/sup>p<sub>r<\/sub><\/p>\n<p>\u2192 The number of permutations of n things taken r at a time containing a particular thing is r \u00d7 <sup>n-1<\/sup>p<sub>r-1<\/sub><\/p>\n<p>\u2192 The number of permutations of n things taken r at a time not containing a particular thing is <sup>n-1<\/sup>p<sub>r<\/sub><\/p>\n<ul>\n<li>The number of permutations of n things taken r at a time allowing repetitions is n<sup>r<\/sup>.<\/li>\n<li>The number of permutations of n things taken not more than r at a time allowing repetitions is \\(\\frac{n\\left(n^{\\mathrm{r}}-1\\right)}{(n-1)}\\)<\/li>\n<\/ul>\n<p>\u2192 The number of permutations of n things of which p things are of one kind and q things are of another kind etc., is \\(\\frac{n !}{p ! q ! \\ldots \\ldots \\cdots}\\)<\/p>\n<p>\u2192 The sum of all possible numbers formed out of all the \u2018n\u2019 digits without zero is (n-1)! (sum of all the digits) (111 &#8230;&#8230;&#8230;.. n times).<\/p>\n<p>\u2192 The sum of all possible numbers formed out of all the \u2018n\u2019 digits which includes zero is [(n -1)! (sum of all the digits) (111 &#8230;&#8230;&#8230;&#8230;&#8230; n times)] &#8211; [( n &#8211; 2)! (sum of all the digits)(111 &#8230;&#8230;&#8230;&#8230;.. (n -1) times]<\/p>\n<p>\u2192 The sum of all possible numbers formed by taking r digits from the given n digits which do not include zero is <sup>n-1<\/sup>p<sub>r-1<\/sub> (sum of all the digits)(111 &#8230;&#8230;&#8230;&#8230;&#8230; r times).<\/p>\n<p>\u2192 The sum of all possible numbers formed by taking r digits from the given n digits which include zero is <sup>n-1<\/sup>p<sub>r-1<\/sub>(sum of all the digits)(111 &#8230;&#8230;&#8230;&#8230;&#8230;&#8230;.. r times) &#8211; <sup>n-2<\/sup>p<sub>r-2<\/sub>(sum of all the digits)(111 &#8230;&#8230;&#8230;&#8230;&#8230;&#8230;.. (r-1) times).<\/p>\n<ul>\n<li>The number of permutations of n things when arranged round a circle is (n -1)!<\/li>\n<li>In case of necklace or garland number of circular permutations is \\(\\frac{(n-1) !}{2}\\)<\/li>\n<\/ul>\n<p><img loading=\"lazy\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/07\/AP-Board-Solutions.png\" alt=\"Inter 2nd Year Maths 2A Permutations and Combinations Formulas\" width=\"196\" height=\"17\" \/><\/p>\n<p>\u2192 Number of permutations of n things taken r at a time in which there is at least one repetition is n<sup>r<\/sup> &#8211; <sup>n<\/sup>p<sub>r<\/sub>.<\/p>\n<p>\u2192 The number of circular permutations of \u2018n\u2019 different things taken \u2018r\u2019 at a time is \\(\\frac{{ }^{n} \\mathrm{P}_{\\mathrm{r}}}{\\mathrm{r}}\\)<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Use these Inter 2nd Year Maths 2A Formulas PDF Chapter 5 Permutations and Combinations to solve questions creatively. Intermediate 2nd Year Maths 2A Permutations and Combinations Formulas \u2192 Fundamental principle: If a work can be performed in m different ways and another work can be performed in n different ways, then the two works simultaneously &#8230; <a title=\"Inter 2nd Year Maths 2A Permutations and Combinations Formulas\" class=\"read-more\" href=\"https:\/\/apboardsolutions.com\/inter-2nd-year-maths-2a-permutations-and-combinations-formulas\/\" aria-label=\"More on Inter 2nd Year Maths 2A Permutations and Combinations Formulas\">Read more<\/a><\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[14],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/posts\/19867"}],"collection":[{"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/comments?post=19867"}],"version-history":[{"count":8,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/posts\/19867\/revisions"}],"predecessor-version":[{"id":20444,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/posts\/19867\/revisions\/20444"}],"wp:attachment":[{"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/media?parent=19867"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/categories?post=19867"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/tags?post=19867"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}