Advertisements
Advertisements
प्रश्न
Find n if: `("n"!)/(3!("n" - 3)!) : ("n"!)/(5!("n" - 5)!)` = 5:3
उत्तर
`("n"!)/(3!("n" - 3)!) : ("n"!)/(5!("n" - 5)!)` = 5:3
∴ `("n"!)/(3!("n" - 3)!)xx(5!("n"-5)!)/("n"!) = 5/3`
`∴("n"!)/(3!("n"-3)("n"-4)("n"-5)!) xx(5xx4xx3!("n"-5)!)/"n!"=5/3`
∴ `(5xx4)/(("n" - 3)("n" - 4)) = 5/3`
∴ (n – 3) (n – 4) = `(20 xx 3)/(5)`
∴ (n – 3) (n – 4) = 12
∴ (n – 3) (n – 4) = 4 × 3
Comparing on both sides, we get
∴ n – 3 = 4
∴ n = 7
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 – 6)!
Compute: `(6! - 4!)/(4!)`
Compute: `(8!)/(6! - 4!)`
Compute: `(8!)/((6 - 4)!)`
Write in terms of factorial:
3 × 6 × 9 × 12 × 15
Write in terms of factorial:
5 × 10 × 15 × 20 × 25
Find n, if `"n"/(8!) = 3/(6!) + 1/(4!)`
Find n, if `"n"/(6!) = 4/(8!) + 3/(6!)`
Find n, if `1/("n"!) = 1/(4!) - 4/(5!)`
Find n, if (n + 1)! = 42 × (n – 1)!
Find n, if: `((2"n")!)/(7!(2"n" - 7)!) : ("n"!)/(4!("n" - 4)!)` = 24:1
Show that
`("n"!)/("r"!("n" - "r")!) + ("n"!)/(("r" - 1)!("n" - "r" + 1)!) = (("n" + 1)!)/("r"!("n" - "r" + 1)!`
Find the value of: `(8! + 5(4!))/(4! - 12)`
Show that: `((2"n")!)/("n"!)` = 2n(2n – 1)(2n – 3)....5.3.1
Five balls are to be placed in three boxes, where each box can contain up to five balls. Find the number of ways if no box is to remain empty.
A hall has 12 lamps and every lamp can be switched on independently. Find the number of ways of illuminating the hall.
A question paper has 6 questions. How many ways does a student have if he wants to solve at least one question?