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Exercise 6.3 · Q2

Q.Find rr if

(i) 9Pr=3024^9P_r = 3024
(ii) 5Pr=2 6Pr−1^5P_r = 2\ ^6P_{r-1}
Sikkim CbseNCERTSubjective· 3mImportance★★★★★est
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✓ Free question

Use nPr=n!(n−r)!^nP_r=\dfrac{n!}{(n-r)!} and solve for rr in each equation.

nPr=n!(n−r)!^nP_r=\dfrac{n!}{(n-r)!}, where nn = total items, rr = items chosen/arranged (0≤r≤n0\le r\le n).

(i) 9Pr=3024^9P_r = 3024

  1. Compute successive values of 9Pr^9P_r: 9P1=9, 9P2=9×8=72, 9P3=72×7=504, 9P4=504×6=3024^9P_1=9,\ ^9P_2=9\times8=72,\ ^9P_3=72\times7=504,\ ^9P_4=504\times6=3024.
  2. Since 9P4=3024^9P_4=3024 matches, r=4r=4.
  3. Check: 9P4=9!5!=9×8×7×6=3024^9P_4=\dfrac{9!}{5!}=9\times8\times7\times6=3024. ✓

(ii) 5Pr=2⋅6Pr−1^5P_r = 2\cdot{}^6P_{r-1}

  1. Write both sides using the formula: 5!(5−r)!=2⋅6!(6−(r−1))!=2⋅6!(7−r)!\dfrac{5!}{(5-r)!} = 2\cdot\dfrac{6!}{(6-(r-1))!} = 2\cdot\dfrac{6!}{(7-r)!}.
  2. Cross-multiply: (7−r)!(5−r)!=2⋅6!5!=2×6=12\dfrac{(7-r)!}{(5-r)!} = \dfrac{2\cdot6!}{5!} = 2\times6=12.
  3. (7−r)!/(5−r)!=(7−r)(6−r)(7-r)!/(5-r)! = (7-r)(6-r), so (7−r)(6−r)=12(7-r)(6-r)=12.
  4. Expand: 42−13r+r2=12⇒r2−13r+30=042-13r+r^2=12 \Rightarrow r^2-13r+30=0.
  5. Factor: (r−3)(r−10)=0⇒r=3(r-3)(r-10)=0 \Rightarrow r=3 or r=10r=10.
  6. Since 5Pr^5P_r requires r≤5r\le5, reject r=10r=10; so r=3r=3.
  7. Check: 5P3=5×4×3=60^5P_3=5\times4\times3=60; 6P2=6×5=30^6P_2=6\times5=30; 2×30=602\times30=60. ✓
✓Final answer

(i) r=4r=4 (ii) r=3r=3

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