Permutations when all objects are distinct [r ≤ n]

This sub-section proves Theorem 1: the number of permutations of $n$ distinct objects taken $r$ at a time (with $r\le n$), without repetitions, is $n\times(n-1)\times(n-2)\times\cdots\times(n-r+1)$.

Proof. Arranging $n$ distinct objects taken $r$ at a time without repetitions is the same problem as filling $r$ places in a row using $n$ available objects (with $r\le n$). Consider a table listing each place from 1st to $r$th across the top, and the number of ways available to fill each place across the bottom: the 1st place can be filled with any of the $n$ objects, so there are $n$ ways. After placing one object there, only $n-1$ objects remain (since repetition is not allowed), so the 2nd place can be filled in $n-1$ ways. After two places are filled, only $n-2$ objects remain for the 3rd place, giving $n-2$ ways. Continuing this pattern, after $r-1$ places have been filled, only $n-(r-1)$ objects remain available, so the $r$th place can be filled in $[n-(r-1)]$ ways. By the Multiplication Principle, the total number of ways to fill all $r$ places is the product of these $r$ shrinking counts: ${}^nP_r = n\times(n-1)\times(n-2)\times\cdots\times(n-r+2)\times(n-r+1).$

This product can be converted into a clean factorial ratio by a standard trick: multiply and divide by the SAME quantity, $(n-r)\times(n-r-1)\times\cdots\times3\times2\times1=(n-r)!$ (the product of all the integers that were 'skipped' below $n-r+1$). Doing so, the numerator $n\times(n-1)\times\cdots\times(n-r+1)\times(n-r)\times(n-r-1)\times\cdots\times1$ becomes exactly $n!$ (every integer from 1 to $n$, once each), while the newly-introduced denominator is $(n-r)!$. This gives the closed formula: ${}^nP_r = \dfrac{n!}{(n-r)!} \quad \text{(for } r\le n\text{)}.$

Note (the special case $r=n$). When all $n$ objects are placed in a row ($r=n$), direct substitution into the shrinking-product form gives ${}^nP_n = n\times(n-1)\times(n-2)\times\cdots\times[n-(n-1)] = n\times(n-1)\times\cdots\times1 = n!.$ This can also be recovered directly from the closed formula: ${}^nP_n = \dfrac{n!}{(n-n)!}=\dfrac{n!}{0!}=\dfrac{n!}{1}=n!$ since $0!=1$ by the defining convention of §3.4 — confirming that the general formula and the direct special-case reasoning agree exactly, which is precisely why the $0!=1$ convention was chosen in the first place.

Solved Example 1. Find the value of ${}^5P_2$. Using the formula, ${}^5P_2 = \dfrac{5!}{(5-2)!} = \dfrac{5!}{3!} = \dfrac{5\times4\times3!}{3!} = 5\times4=20.$

Solved Example 2. How many different ways are there to arrange the letters of the word WORLD? How many of these begin with R? How many can be made taking three letters at a time? WORLD has 5 distinct letters, so all of them arranged among themselves gives ${}^5P_5=5!=120$ different arrangements. If an arrangement must BEGIN with R, the letter R is fixed in the first position, and the remaining 4 distinct letters (W,O,L,D) are arranged among themselves in the remaining 4 positions: ${}^4P_4=4!=24$ ways. Finally, the number of arrangements of the 5 letters taken 3 at a time is ${}^5P_3=\dfrac{5!}{2!}=5\times4\times3=60$.

Solved Example 3. How many three-digit numbers can be formed from the digits 2,4,5,6,7 if no digit is repeated? Every distinct arrangement of two of these five digits (into the tens and units-equivalent roles alongside a leading digit choice) gives a different number, so the problem reduces to finding the number of arrangements of 5 digits taken 3 at a time (none of the digits is 0, so there is no leading-digit restriction to worry about): ${}^5P_3 = \dfrac{5!}{2!} = \dfrac{5\times4\times3\times2!}{2!}=5\times4\times3=60.$

Solved Example 4. How many numbers can be formed with the digits 3,4,6,7,8 taken all at a time, and what is the sum of all such numbers? The five distinct digits, arranged all at a time, give ${}^5P_5=5!=120$ different five-digit numbers. To find their sum, focus on a single digit, say 3, and ask: in how many of the 120 numbers does 3 appear in the UNIT's place specifically? If 3 is fixed in the unit's place, the other 4 digits can be arranged in the remaining 4 positions in ${}^4P_4=4!=24$ ways — so 3 appears in the unit's place in exactly 24 of the 120 numbers. By the same reasoning, each of the other four digits (4, 6, 7, 8) also appears in the unit's place in exactly 24 of the 120 numbers. So the sum of just the UNIT's-place digits, across all 120 numbers, is $24\times(3+4+6+7+8)=24\times28=672$. By an identical argument, the same total, 672, is contributed by the digits appearing in the ten's place across all 120 numbers, and likewise for the hundred's, thousand's, and ten-thousand's places (each digit is equally likely, by symmetry, to occupy any of the 5 positions across the full set of arrangements). So the grand total sum of all 120 numbers is $672\times(1+10+100+1000+10000) = 672\times11111 = 74,66,592$. …