Question

ज्ञात कीजिए, यदि (i) $^{5}P_r = 2 \,^{6}P_{r-1}$ (ii) $^{5}P_r = \,^{6}P_{r-1}$

Answer 1

Answer:

Step-by-step explanation:

(i)  प्रश्नानुसार  

         $^{5}P_r = 2 \,^{6}P_{r-1}\\\\\frac{5!}{(5-r)!} =2[\frac{6!}{(6-r+1)!} ]\\\\\frac{5!}{(5-r)!} =2[\frac{6*5!}{(7-r)!} ]\\\\\frac{1}{(5-r)!} =\frac{12}{(7-r)(6-r)(5-r)!} \\\\(7-r)(6-r)=12\\\\42-13r+r^2=12\\\\r^2-13r+30=0\\\\(r-10)(r-3)=0$

  r =  10  या   3

अतः   r = 3  क्योंकि  r = 10 अर्थहीन है।

(ii)  प्रश्नानुसार  

             $^{5}P_r = \,^{6}P_{r-1}\\\\\frac{5!}{(5-r)!} =\frac{6!}{[6-(r-1)]!} \\\\\frac{5!}{(5-r)!} =\frac{6*5!}{[6-r+1]!} \\\\\frac{5!}{(5-r)!} =\frac{6*5!}{(7-r)!} \\\\\frac{5!}{(5-r)!} =\frac{6*5!}{(7-r)(6-r)(5-r)!} \\\\(7-r)(6-r)=6\\\\r^2-13r+36=0\\\\(r-9)(r-4)=0$

  r  =  9  या  4  

∴   r  =  4 क्योंकि   r ≠ 9 क्योंकि यह 5 से बड़ा है |