We all face problems with P&C. But not anymore.
|
| Forum Index -> Community shelf -> View Full Question |
|
| Author | Message | |||||||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
|
Factorial: the continued product of first n natural numbers is called the “n factorial” and is denoted by n! or i.e. n! = 1*2*3…*(n-1)*n 4! = 1*2*2*4 n! is defined for positive integers only. 0! is defined as 1. n! = n*(n-1)! Exponent of prime number p in factorial of n (n!)The exponent of prime number of 3 in 100! is 48. It means 100! is divisible by 348 How do you find it? Let p be a prime number and n be a positive integer. Then find the last integer in the sequence 1,2,…,n which is divisible by p. Express this integer as [n/p]p. [n/p] denotes the greatest integer less than or equal to n/p In case of 3 (p) and 100 (n); [n/p] is 33 and n/p is 33 and 1/3. Let Ep(n) denote the exponent of the prime p in the positive integer n. Then, Ep(n!) = Ep(1.2.3…(n-1).n) This will be equal to Ep(p.2p.3p…[n/p]p) = [n/p]+ Ep(1.2.3...[n/p]) This process continues and we get the answer Ep(n!) = [n/p] + [n/p²]+…+[n/ps] Where s is the largest positive integer such that ps≤n≤ps+1 Hence applying the formula to find exponent of prime 3 in 100! E3(100!) = [100/3] + [100/3²] + [100/3³] + [100/34] = 33+11+3+1 = 48 Note: remember the meaning of notation [100/3] or [n/p] Fundamental principle of multiplicationIf there are two jobs such that one of them can be completed in m ways, and when it has been completed in any one of these m ways, second job can be completed in n ways, then the two jobs in succession can be completed in m*n ways. Fundamental principle of addition If there are two jobs such that they can be performed independently in m and n ways respectively, then either of the two jobs can be performed in (m+n) ways. Permutations Definition and TheoremsEach of the arrangement which can be made by taking some or all of a number of things is called a permutation. Theorem 1 Let r and n be positive integers such that 1≤r≤n. then the number of all permutations of n distinct things taken r at a time is given by n(n-1)(n-2)…(n-(r-1)) Notation: Let r and n be positive integers such that 1≤r≤n. then the number of all permutations of n distinct things taken r at a time is denoted by the symbol P(n,r) or n Cr. Permutations under certain conditionsThree theorems Theorem 1 The number of all permutations of n different objects taken r at a time, when a particular object is to be always included in each arrangement is r.n-1Cr-1 Theorem 2 The number of all permutations of n different objects taken r at a time, when a particular object is never taken in each arrangement is, n-1Cr-1 Theorem 3 The number of all permutations of n different objects taken r at a time, when two specified objects always occur together is 2!(r-1) n-2Cr-2 Then P(n,r) = n Cr = n(n-1)(n-2)…(n-(r-1)) Permutations of Objects not all DistinctTheorems and Formulas Theorem The number of mutually distinguishable permutations of n things, taken all at a time, of which p are alike of one kind, q alike of second such that p+q = n, is n!/p!q! Formulas based on the above theorem 1. The number of mutually distinguishable permutations of n things, taken all at a time, of which p1 are alike of one kind, p2 alike of second,…, pr alike of of rth kind such that p1+p2+…pr = n, is n!/p1!p2!…pr! 2. The number of mutually distinguishable permutations of n tings, of which p are alike of one kind, q alike of second and remaining all are distinct is n!/p!q! 3. suppose there are r things to be arranged, allowing repetitions. Let further p1,p2,…,pr be the integers such that the first object occurs exactly p1 times, the second occurs exactly p2 times, etc. Then the total number of permutations of these r objects to the above condition is (p1+p2+…+pr)!/p1!p2!…pr! Combinations and Difference between Combinations and PermutationsEach of the different selections made by taking some or all of a number of objects, irrespective of their arrangement is called a combination. Difference between Combinations and Permutations In a combination, the ordering of the selected objects is immaterial whereas in a permutation, the ordering is essential. For example AB and BA are same as combinations, but different as permutations. Associate the word selection for combinations and arrangement for permuations. Notation and Theorem for CombinationsNotation The number of all combinations of n objects, taken r at a time is denoted by C(n,r) or nCr. nCr is defined when n and r are non-negative numbers. Theorem The number of all combinations of n distinct objects, taken r at a time is given by nCr = n!/(n-r)!r! Results from the theorem nCr = [n(n-1)(n-2)...(n-r+1)]/(1.2.3...r) nCn =1 nC0 = 1 nCr = nPr/r! Properties of C(n,r)Properties of nCr or C(n,r) 1. nCr = nCn-r Note: If x=y = n nCx = nCy 2. Let n and r be non-negative integers such that r≤n. Then nCr + nCr-1 = n+1Cr 3. Let n and r be non-negative integers such that 1≤ r≤n. Then nCr = (n/r) n-1Cr-1 4. If 1≤ r≤n, then n.n-1Cr-1 = (n-r+1)nCr-1 5. nCx = nCy implies x+y = n 6. If n is even, then the greatest value of nCr [0≤ r≤n] is nCn/2. 7. If n is odd, then the greatest value of nCr [0≤ r≤n] is nC(n+1)/2 or nC(n-1)/2. Hope u like it. Theorem 2 P(n,r) = n Cr = n!/(n-r)! Theorem 3 The number of all permutations of n distinct things taken all at a time is n!. Theorem 4 0! = 1 |
|||||||||||||
| Like 0 people liked this | ||||||||||||||
|
|
||||||||||||||












