Advertisements
Advertisements
प्रश्न
Find n, if `1/("n"!) = 1/(4!) - 4/(5!)`
Advertisements
उत्तर
`1/("n"!) = 1/(4!) - 4/(5!)`
∴ `1/("n"!) = 1/(4!) - 4/(5!)`
∴ `1/("n"!) = 5/(5xx4!)- 4/(5!)`
∴ `1/("n"!) = 5/(5!)-4/(5!)`
∴ `1/("n"!) = 1/(5!)`
∴ n! = 5!
∴ n = 5
APPEARS IN
संबंधित प्रश्न
If numbers are formed using digits 2, 3, 4, 5, 6 without repetition, how many of them will exceed 400?
Evaluate: 8!
Evaluate: 6!
Evaluate: 8! – 6!
Compute: (3 × 2)!
Compute: 3! × 2!
Compute: `(9!)/(3! 6!)`
Compute: `(8!)/(6! - 4!)`
Write in terms of factorial:
3 × 6 × 9 × 12 × 15
Evaluate: `("n"!)/("r"!("n" - "r"!)` For n = 8, r = 6
Find n, if (n + 1)! = 42 × (n – 1)!
Find n if: `("n"!)/(3!("n" - 5)!) : ("n"!)/(5!("n" - 7)!)` = 10:3
Find n, if: `((15 - "n")!)/((13 - "n")!)` = 12
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 student passes an examination if he/she secures a minimum in each of the 7 subjects. Find the number of ways a student can fail.
A question paper has 6 questions. How many ways does a student have if he wants to solve at least one question?
