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Exercise 6.3 · Q3

Q.Prove the following:

(i) nPn=2.nPn−2^nP_n = 2 . {}^nP_{n-2}
(ii) n−1Pr+r.n−1Pr−1=nPr^{n-1}P_r + r . {}^{n-1}P_{r-1} = {}^nP_r
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✓ Free question

Expand each side with nPr=n!(n−r)!^nP_r=\dfrac{n!}{(n-r)!} and show LHS = RHS.

nPr=n!(n−r)!^nP_r=\dfrac{n!}{(n-r)!}; n!=n×(n−1)×(n−2)×⋯×1n!=n\times(n-1)\times(n-2)\times\cdots\times1.

(i) Prove nPn=2⋅nPn−2^nP_n = 2\cdot{}^nP_{n-2}

  1. nPn=n!(n−n)!=n!0!=n!^nP_n=\dfrac{n!}{(n-n)!}=\dfrac{n!}{0!}=n!.
  2. nPn−2=n!(n−(n−2))!=n!2!=n!2^nP_{n-2}=\dfrac{n!}{(n-(n-2))!}=\dfrac{n!}{2!}=\dfrac{n!}{2}.
  3. So 2⋅nPn−2=2×n!2=n!2\cdot{}^nP_{n-2}=2\times\dfrac{n!}{2}=n!.
  4. LHS =n!==n!= RHS. Hence proved (valid for n≥2n\ge2).

(ii) Prove (n−1)Pr+r⋅(n−1)Pr−1=nPr^{(n-1)}P_r + r\cdot{}^{(n-1)}P_{r-1} = {}^nP_r

  1. (n−1)Pr=(n−1)!(n−1−r)!^{(n-1)}P_r=\dfrac{(n-1)!}{(n-1-r)!} and (n−1)Pr−1=(n−1)!(n−1−(r−1))!=(n−1)!(n−r)!^{(n-1)}P_{r-1}=\dfrac{(n-1)!}{(n-1-(r-1))!}=\dfrac{(n-1)!}{(n-r)!}.
  2. Write (n−r)!=(n−r)(n−r−1)!=(n−r)(n−1−r)!(n-r)!=(n-r)(n-r-1)!=(n-r)(n-1-r)!, so (n−1)Pr−1=(n−1)!(n−r)(n−1−r)!^{(n-1)}P_{r-1}=\dfrac{(n-1)!}{(n-r)(n-1-r)!}.
  3. LHS =(n−1)!(n−1−r)!+r⋅(n−1)!(n−r)(n−1−r)!=(n−1)!(n−1−r)![1+rn−r]=\dfrac{(n-1)!}{(n-1-r)!}+r\cdot\dfrac{(n-1)!}{(n-r)(n-1-r)!} =\dfrac{(n-1)!}{(n-1-r)!}\left[1+\dfrac{r}{n-r}\right].
  4. Combine the bracket: 1+rn−r=(n−r)+rn−r=nn−r1+\dfrac{r}{n-r}=\dfrac{(n-r)+r}{n-r}=\dfrac{n}{n-r}.
  5. So LHS =(n−1)!(n−1−r)!×nn−r=n⋅(n−1)!(n−r)(n−1−r)!=n!(n−r)!=\dfrac{(n-1)!}{(n-1-r)!}\times\dfrac{n}{n-r} = \dfrac{n\cdot(n-1)!}{(n-r)(n-1-r)!}=\dfrac{n!}{(n-r)!} (since (n−r)(n−1−r)!=(n−r)!(n-r)(n-1-r)!=(n-r)!).
  6. This equals nPr^nP_r by definition. Hence LHS == RHS, proved.
✓Final answer

Both identities hold identically for all valid n,rn,r — proofs complete as shown.

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