ⁿP₄=14×^n-2P₃ হলে n=?

Explanation:

Another Explanation (5): bài toán: ⁿP₄=14×^n-2P₃ হলে n=? 🤔 আমরা জানি, ⁿP_r = \(\frac{n!}{(n-r)!}\) 🤓 তাহলে, ⁿP₄ = \(\frac{n!}{(n-4)!}\) এবং ^n-2P₃ = \(\frac{(n-2)!}{(n-2-3)!}\) = \(\frac{(n-2)!}{(n-5)!}\) 😎 সুতরাং, \(\frac{n!}{(n-4)!}\) = 14 × \(\frac{(n-2)!}{(n-5)!}\) 🥳 \(\implies\) \(\frac{n(n-1)(n-2)(n-3)(n-4)!}{(n-4)!}\) = 14 × \(\frac{(n-2)(n-3)(n-4)(n-5)!}{(n-5)!}\) 🤩 \(\implies\) n(n-1)(n-2)(n-3) = 14(n-2)(n-3)(n-4) 🤫 \(\implies\) n(n-1) = 14(n-4) [যেহেতু, (n-2)(n-3) ≠ 0] 🤗 \(\implies\) n² - n = 14n - 56 🥰 \(\implies\) n² - 15n + 56 = 0 🤯 \(\implies\) n² - 7n - 8n + 56 = 0 😵‍💫 \(\implies\) n(n-7) - 8(n-7) = 0 🫠 \(\implies\) (n-7)(n-8) = 0 💀 সুতরাং, n = 7 অথবা n = 8। 🤓