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| Permutations |
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| The different arrangements that can be made with a given number of things taking some or all of them at a time are called permutations. |
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| The symbol nPr or P(n,r) is used to denote the number of permutations of n things taken r at a time. |
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| Examples: |
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| 1. 2 and 3 are two digits and with these digits, the numbers 32 and 23 are formed. Although both numbers viz., 32 and 23 consist of the digits 2 and 3, the order of digits is different. Each of the above arrangements is called a 'permutation'. Thus, the number of arrangements or permutations of two distinct digits 2 and 3 is 2. |
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2. The permutation of the three letters a, b, c taken two at a time are  |
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The number of permutations of n dissimilar things taken r at  |
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| The number of permutations of n different things taken r at a time is the same as the number of ways of filling n letters and r blank spaces, supposed to be arranged in a straight line as shown above. Each blank is accommodating only one letter. We may fill the first blank with any one of the n letters. Therefore, the first blank can be filled in n different ways. Having filled the first blank in any one of these ways, we have |
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| (n-1) letters with which to fill the next blank. Hence, the second blank can be filled in (n-1) ways, after filling the first blank in n different ways. |
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| Therefore, there are n(n-1) different ways of filling the first two blank spaces. |
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| Having filled the first two blank spaces in any one of these ways, we have (n-2) letters with which to fill the third blank space, for each of the n(n-1) ways of filling the first two. |
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 There are n(n-1)(n-2) different ways of filling the first three blank spaces. Proceeding in this way, we see that there are n(n-1)(n-2)(n-3)(n-4)....[r factors] different ways of filling r blank spaces with n letters. The rth factor = n - (r -1) = n - r = 1. |
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 The number of r-permutations of n different things is nPr = P(n,r) = n(n-1) (n-2) (n-3)...(n - r +1) |
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| Corollary 1: |
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| If we put r = n in the above formula, then |
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| Note: |
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| Strictly speaking 0! has no meaning. But since nPn = n! we may understand 0! = 1. |
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| Corollary 2: |
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| Corollary 3: |
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| Corollary 4: |
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| First method: |
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| P(n,r) is the number of permutations of n dissimilar things taken r at a time. These permutations can be divided into two groups. |
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| (i) Those not containing a particular thing l. |
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| (ii) Those containing a particular thing l. |
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| Taking out l from the given things, we have (n-1) things which can be arranged in r blanks in P(n-1,r) ways. Then, the number of permutations not containing l is P(n-1,r). |
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| Next in the permutations containing l, it occupies one of the r blank spaces in r ways. The remaining (r-1) blank spaces can be filled with the remaining (n-1) things in P(n-1,r-1) ways. Thus, the number of permutations containing l is lP(n-1, r-1). |
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| Hence, the total number of permutations is P(n-1,r) + rP (n-1, r-1). |
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| Second method: |
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