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Permutations and Combinations question

2020 · 3 Sep · Shift 1 · Q26
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Permutations and Combinations question

2020 · 3 Sep · Shift 1 · Q26

JEE MainMathematicsPermutations and CombinationsMCQ+4 / −1
The value of (2.1P0 – 3.2P1 + 4.3P2 .... up to 51th term) + (1! – 2! + 3! – ..... up to 51th term) is equal to :
  1. A
    1
  2. B
    1 + (51)!
  3. C
    1 – 51(51)!
  4. D
    1 + (52)!
View written solutionFree

Correct answer: D

  1. Interpret the first series carefully

The pattern is 2⋅1P0−3⋅2P1+4⋅3P2−⋯2\cdot {}^1P_0-3\cdot {}^2P_1+4\cdot {}^3P_2-\cdots2⋅1P0​−3⋅2P1​+4⋅3P2​−⋯ up to the 51st51^{\text{st}}51st term.

So the general term is (−1)r−1(r+1)⋅rPr−1,r=1,2,3,…,51(-1)^{r-1}(r+1)\cdot {}^rP_{r-1}, \quad r=1,2,3,\dots,51(−1)r−1(r+1)⋅rPr−1​,r=1,2,3,…,51

because:

  • for r=1r=1r=1: 2⋅1P02\cdot {}^1P_02⋅1P0​
  • for r=2r=2r=2: −3⋅2P1-3\cdot {}^2P_1−3⋅2P1​
  • for r=3r=3r=3: +4⋅3P2+4\cdot {}^3P_2+4⋅3P2​
  • etc.

Now, rPr−1=r!(r−(r−1))!=r!1!=r!{}^rP_{r-1}=\frac{r!}{(r-(r-1))!}=\frac{r!}{1!}=r!rPr−1​=(r−(r−1))!r!​=1!r!​=r!

Hence the first sum becomes S1=∑r=151(−1)r−1(r+1)r!S_1=\sum_{r=1}^{51}(-1)^{r-1}(r+1)r!S1​=∑r=151​(−1)r−1(r+1)r!

But (r+1)r!=(r+1)!(r+1)r!=(r+1)!(r+1)r!=(r+1)!

So, S1=∑r=151(−1)r−1(r+1)!S_1=\sum_{r=1}^{51}(-1)^{r-1}(r+1)!S1​=∑r=151​(−1)r−1(r+1)!

That is, S1=2!−3!+4!−5!+⋯+52!S_1=2!-3!+4!-5!+\cdots+52!S1​=2!−3!+4!−5!+⋯+52!

  1. Write the second series

The second sum is S2=1!−2!+3!−4!+⋯+51!S_2=1!-2!+3!-4!+\cdots+51!S2​=1!−2!+3!−4!+⋯+51!

In sigma form, S2=∑r=151(−1)r−1r!S_2=\sum_{r=1}^{51}(-1)^{r-1}r!S2​=∑r=151​(−1)r−1r!

  1. Add the two sums termwise

We need S=S1+S2S=S_1+S_2S=S1​+S2​

So, S=(2!−3!+4!−5!+⋯+52!)+(1!−2!+3!−4!+⋯+51!)S=(2!-3!+4!-5!+\cdots+52!)+(1!-2!+3!-4!+\cdots+51!)S=(2!−3!+4!−5!+⋯+52!)+(1!−2!+3!−4!+⋯+51!)

Now combine like factorial terms:

  • 1!1!1! appears only once: gives 1!1!1!
  • 2!2!2! cancels: +2!−2!=0+2!-2!=0+2!−2!=0
  • 3!3!3! cancels: −3!+3!=0-3!+3!=0−3!+3!=0
  • 4!4!4! cancels: +4!−4!=0+4!-4!=0+4!−4!=0
  • and so on...
  • 51!51!51! cancels
  • 52!52!52! remains

Therefore, S=1!+52!=1+52!S=1!+52!=1+52!S=1!+52!=1+52!

  1. Match with the options

Thus the value is 1+52!\boxed{1+52!}1+52!​

So the correct option is:

D\boxed{\text{D}}D​

  1. Comparison with stored correct answer

Stored correct answer: D

Our derived answer: D

They agree.

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