The value of $\left(2 \cdot {}^{1}P_{0}-3 \cdot {}^{2}P_{1}+4 \cdot {}^{3}P_{2}-\ldots\right.$ up to $51^{\text{th}}$ term$)+\left(1!-2!+3!-\ldots\right.$ upt to $51^{\text{th}}$ term) is equal to:

JEE Main · 2020 · 03 Sep 2020 · Shift 1 · Permutations and Combinations

The value of

(2 \cdot^{1} P_{0} - 3 \cdot^{2} P_{1} + 4 \cdot^{3} P_{2} - \dots

up to

5 1^{th}

term

) + (1! - 2! + 3! - \dots

upt to

5 1^{th}

term) is equal to:

Answer: (C)

Because

^{k} P_{k - 1} = k!

, the first bracket is

2 \cdot 1! - 3 \cdot 2! + 4 \cdot 3! - \dots + 52 \cdot 51!

, and each term

(r + 1) \cdot r! = (r + 1)!

, so it equals

2! - 3! + 4! - \dots + 52!

. The second bracket is

1! - 2! + 3! - \dots + 51!

. Adding the two, every factorial from

2!

up to

51!

appears once with each sign and cancels, leaving only

1!

from the second bracket and

52!

from the first, so the value is

1 + (52)!

JEE Main · 2020 · 03 Sep 2020 · Shift 1 · Single correct option