Q.Find n, if ${}^nP_6 : {}^nP_3 = 120:1$

When all $n$ objects available are distinct (no two are alike) and repetition is NOT allowed, the number of ways to arrange $r$ of them (where $r\le n$) in a definite order is denoted ${}^nP_r$ and is given by ${}^nP_r = n\times(n-1)\times(n-2)\times\cdots\times(n-r+1) = \dfrac{n!}{(n-r)!}$. The reasoning behind the formula is a direct application of the Multiplication Principle: the first of the $r$ positions can be filled in $n$ ways (any of the $n$ objects); having used one object, the second position can be filled in only $n-1$ ways (one fewer object remains); the third in $n-2$ ways; and so on, until the $r$-th position, which can be filled in $n-(r-1)$ ways since $r-1$ objects have already been used. Multiplying these $r$ shrinking counts together gives the formula. A special case worth remembering is $r=n$: arranging ALL $n$ distinct objects gives ${}^nP_n = n!$ (using the convention $0!=1$). This formula, and the theorems built on top of it — such as the count when a specified object must always (or must never) appear, or when a fixed group of objects must stay together — is the workhorse for problems about forming numbers from digit sets, arranging distinguishable people or books in a row, and forming words from sets of distinct letters, always under the constraint that no object may be used more than once and the arrangement is in a straight line (not a circle).