COMPASSLearning

Talk to us Register Login

JEE Main · 2021 · 26 Aug 2021 · Shift 1 · Permutations and Combinations

If

^{1} P_{1} + 2 \cdot^{2} P_{2} + 3 \cdot^{3} P_{3} + \dots\dots\dots + 15 \cdot^{15} P_{15} =^{q} P_{r} - s, 0 \leq s \leq 1

, then

^{q + s} C_{r - s}

is equal to ______.

Answer: 136

Solution

Since

^{r} P_{r} = r!

, each term is

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

, so the sum telescopes:

r = 1 \sum 15 r \cdot^{r} P_{r} = r = 1 \sum 15 [(r + 1)! - r!] = 16! - 1! = 16! - 1.

Comparing with

^{q} P_{r} - s

under

0 \leq s \leq 1

gives

s = 1

and

^{q} P_{r} = 16! =^{16} P_{16}

, so

q = 16

and

r = 16

.
Hence

^{q + s} C_{r - s} =^{17} C_{15} = \frac{17 \cdot 16}{2} = 136.

JEE Main · 2021 · 26 Aug 2021 · Shift 1 · Integer answer

Practise more: all JEE Main PYQs on Permutations and Combinations →

More JEE Main questions on Permutations and Combinations

The number of strictly increasing functions f from the set ${1, 2, 3, 4, 5, 6}$ to the set $(1, 2, 3, \dots, 9)$ such that $f (i) \neq = i$ for $1 \leq i \leq 6$ , is equal to :
JEE Main 2026View solution →
Let $S = {(m, n) : m, n \in {1, 2, 3, \dots .., 50}}$ . If the number of elements ( $m, n$ ) in S such that $6^{m} + 9^{n}$ is a multiple of 5 is p and the number of elements …
JEE Main 2026View solution →
The largest $n \in N$ , for which $7^{n}$ divides 101!, is:
JEE Main 2026View solution →
Let ABC be a triangle. Consider four points $p_{1}, p_{2}$ , $p_{3}, p_{4}$ on the side AB , five points $p_{5}, p_{6}, p_{7}, p_{8}, p_{9}$ on the side BC and four points …
JEE Main 2026View solution →
Let S be the set of the first 11 natural numbers. Then the number of elements in $A = {B \subseteq S : n (B) \geq 2$ and the product of all elements of B is even $}$ is
JEE Main 2026View solution →
The number of 4-letter words, with or without meaning, which can be formed using the letters PQRPQRSTUVP, is ______.
JEE Main 2026View solution →