Advertisements
Advertisements
प्रश्न
Find n, if: `((17 - "n")!)/((14 - "n")!)` = 5!
Advertisements
उत्तर
`((17 - "n")!)/((14 - "n")!)` = 5!
∴ `((17 - "n")(16 - "n")(15 - "n")(14 - "n")!)/((14 - "n")!)` = 5 × 4 × 3 × 2 × 1
∴ (17 – n) (16 – n) (15 – n) = 6 × 5 × 4
Comparing on both sides, we get
17 – n = 6
∴ n = 11
APPEARS IN
संबंधित प्रश्न
If numbers are formed using digits 2, 3, 4, 5, 6 without repetition, how many of them will exceed 400?
How many five-digit numbers formed using the digit 0, 1, 2, 3, 4, 5 are divisible by 3 if digits are not repeated?
Evaluate: 8! – 6!
Evaluate: (8 – 6)!
Compute: `(12/6)!`
Compute: `(8!)/(6! - 4!)`
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"/(6!) = 4/(8!) + 3/(6!)`
Find n, if (n + 1)! = 42 × (n – 1)!
Find n if: `("n"!)/(3!("n" - 3)!) : ("n"!)/(5!("n" - 5)!)` = 5:3
Find n if: `("n"!)/(3!("n" - 5)!) : ("n"!)/(5!("n" - 7)!)` = 10:3
Show that
`("n"!)/("r"!("n" - "r")!) + ("n"!)/(("r" - 1)!("n" - "r" + 1)!) = (("n" + 1)!)/("r"!("n" - "r" + 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.
How many quadratic equations can be formed using numbers from 0, 2, 4, 5 as coefficient if a coefficient can be repeated in an equation.
A question paper has 6 questions. How many ways does a student have if he wants to solve at least one question?
