Let n be the natural number. Then range of the function f(n) = ^8-nP_n-4 , 4 ≤ n ≤6 is

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$In,^{8 - n} P_{n - 4} It is given that 4 \leq n \leq 6 So at n = 4^{8 - n} P_{n - 4} =^{8 - 4} P_{4 - 4} =^{4} P_{0} = 1 and at n = 5^{8 - n} P_{n - 4} =^{8 - 5} P_{5 - 4} =^{3} P_{1} = 3 and at n = 6^{8 - n} P_{n - 4} =^{8 - 6} P_{6 - 4} =^{2} P_{2} = 2 So range is [1, 3]$

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Let n be the natural number. Then range of the function f(n) = 8-nPn-4 , 4 ≤ n ≤6 is

In, 8-nPn-4It is given that 4≤n≤6So at n = 48-nPn-4=8-4P4-4=4P0=1and at n = 58-nPn-4=8-5P5-4=3P1=3and at n = 68-nPn-4=8-6P6-4=2P2=2So range is 1,3

Let n be the natural number. Then range of the function f(n) = ^8-nP_n-4 , 4 ≤ n ≤6 is

$In,^{8 - n} P_{n - 4} It is given that 4 \leq n \leq 6 So at n = 4^{8 - n} P_{n - 4} =^{8 - 4} P_{4 - 4} =^{4} P_{0} = 1 and at n = 5^{8 - n} P_{n - 4} =^{8 - 5} P_{5 - 4} =^{3} P_{1} = 3 and at n = 6^{8 - n} P_{n - 4} =^{8 - 6} P_{6 - 4} =^{2} P_{2} = 2 So range is [1, 3]$