text: # If $$\dfrac{(x+2)!}{(2x-1)!}\times\dfrac{(2x+1)!}{(x+3)!}=\dfrac{72}{7}$$, where x ∈ N then find the value of x?
  
  This question was previously asked in
  DSSSB Patwari Question Paper 16th June 2019 Shift 1
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  1. $$\dfrac{-54}{28}$$
  2. 4
  3. -4
  4. $$\dfrac{54}{28}$$
  
  Option 2 : 4
  Free
  DSSSB Patwari Mental Ability Sectional Test 1
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  20 Questions 20 Marks 14 Mins
  
  ## Detailed Solution
  
  $$\frac{{(x + 2)!}}{{(2x - 1)!}} \times \frac{{(2x + 1)!}}{{(x + 3)!}} = \frac{{72}}{7}$$
  
  $$\frac{{(x + 2)!}}{{(2x - 1)!}} \times$$$$\frac{{(2x + 1)(2x)(2x - 1)!}}{{(x + 3)(x + 2)!}} = \frac{{72}}{7}$$
  
  $$= \frac{{2x(2x + 1)}}{{x + 3}} = \frac{{72}}{7}$$
  
  $$= \frac{{x(2x + 1)}}{{x + 3}} = \frac{{36}}{7}$$
  
  = 14x2 + 7x = 36x + 108
  
  = 14x2 - 29x - 108 = 0
  
  $$x = \frac{{ - ( - 29) \pm \sqrt {{{( - 29)}^2} + 4 \times 14 \times 108} }}{{2 \times 14}}$$
  
  $$= \frac{{29 \pm 83}}{{28}}$$
  
  $$x = \frac{{29 + 83}}{{28}}$$$$x = \frac{{29 - 83}}{{28}}$$
  
  x = 4, -1.92
  
  Given x ϵ N, N → Natural number
  
  x = 4
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