Advertisements
Advertisements
प्रश्न
Find n, if (n + 3)! = 110 × (n + 1)!
Advertisements
उत्तर
(n + 3)! = 110 × (n + 1)!
∴ (n + 3)(n + 2)(n + 1)! = 110 × (n + 1)!
∴ (n + 3)(n + 2) = 110
∴ n2 + 5n + 6 = 110
∴ n2 + 5n + 6 − 110 = 0
∴ n2 + 5n − 104 = 0
∴ n2 + 13n − 8n − 104 = 0
∴ n(n + 13) − 8(n + 13) = 0
∴ (n + 13)(n − 8) = 0
∴ n + 13 = 0 or n − 8 = 0
∴ n = − 13 or n = 8
But n ∈ N
∴ n ≠ − 13
Hence, n = 8.
APPEARS IN
संबंधित प्रश्न
A teacher wants to select the class monitor in a class of 30 boys and 20 girls. In how many ways can he select a student if the monitor can be a boy or a girl?
How many five-digit numbers formed using the digit 0, 1, 2, 3, 4, 5 are divisible by 3 if digits are not repeated?
Compute: `(12!)/(6!)`
Compute: `(12/6)!`
Compute: 3! × 2!
Compute: `(8!)/(6! - 4!)`
Write in terms of factorial:
5 × 6 × 7 × 8 × 9 × 10
Write in terms of factorial:
3 × 6 × 9 × 12 × 15
Write in terms of factorial:
6 × 7 × 8 × 9
Evaluate: `("n"!)/("r"!("n" - "r"!)` For n = 8, r = 6
Evaluate: `("n"!)/("r"!("n" - "r"!)` For n = 12, r = 12
Find n, if `"n"/(8!) = 3/(6!) + 1/(4!)`
Find n if: `("n"!)/(3!("n" - 5)!) : ("n"!)/(5!("n" - 7)!)` = 10:3
Find n, if: `((2"n")!)/(7!(2"n" - 7)!) : ("n"!)/(4!("n" - 4)!)` = 24:1
Show that: `((2"n")!)/("n"!)` = 2n(2n – 1)(2n – 3)....5.3.1
A question paper has 6 questions. How many ways does a student have if he wants to solve at least one question?
