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Question

Question: Find the value of \({}^8{P_7},{}^{25}{P_5},{}^{24}{P_4},{}^{19}{P_{14}}.\)...

Find the value of 8P7,25P5,24P4,19P14.{}^8{P_7},{}^{25}{P_5},{}^{24}{P_4},{}^{19}{P_{14}}.

Explanation

Solution

Hint: In this question we need to find values of the given permutation values. We will use the formula to compute the permutation, that is nPr=n!(nr)!{}^n{P_r} = \dfrac{{n!}}{{\left( {n - r} \right)!}}. This will help us to solve the expressions.

Complete step-by-step answer:
For 8P7{}^8{P_7}, to solve this we will use the formula nPr=n!(nr)!{}^n{P_r} = \dfrac{{n!}}{{\left( {n - r} \right)!}}.
Using the formula, we get 8P7=8!(87)!=8!{}^8{P_7} = \dfrac{{8!}}{{\left( {8 - 7} \right)!}} = 8!
8P7=40320\Rightarrow {}^8{P_7} = 40320
Similarly, for 25P5{}^{25}{P_5} we get,
25P5=25!(255)!=25!20!=25×24×23×22×21{}^{25}{P_5} = \dfrac{{25!}}{{\left( {25 - 5} \right)!}} = \dfrac{{25!}}{{20!}} = 25 \times 24 \times 23 \times 22 \times 21
25P5=6375600\Rightarrow {}^{25}{P_5} = 6375600
Similarly, for 24P4{}^{24}{P_4} we get,
24P4=24!(244)!=24!20!=24×23×22×21{}^{24}{P_4} = \dfrac{{24!}}{{\left( {24 - 4} \right)!}} = \dfrac{{24!}}{{20!}} = 24 \times 23 \times 22 \times 21
24P4=255024\Rightarrow {}^{24}{P_4} = 255024
Similarly, for 19P14{}^{19}{P_{14}} we get,
19P14=19!(1914)!=19!5!{}^{19}{P_{14}} = \dfrac{{19!}}{{\left( {19 - 14} \right)!}} = \dfrac{{19!}}{{5!}}
19P14==19!5!\Rightarrow {}^{19}{P_{14}} = = \dfrac{{19!}}{{5!}}

Note: Whenever we face such types of problems the value point to remember is that we need to have a good grasp over permutations and its formulas. The most basic formula to calculate permutations has been discussed above and used to solve the given question. However, we must remember that we don’t need to calculate the value of factorial if it is very large.