Advertisements
Advertisements
Question
In how many ways can 10 examination papers be arranged so that the best and the worst papers never come together?
Options
9 × 8!
8 × 8!
9 × 9!
8 × 9!
Advertisements
Solution
8 × 9!
Explanation:
Arrange 8 papers in 8! ways and two papers in 9 gaps are arranged in 9P2 ways.
Required number = 8! 9P2
= 8! × 9 × 8
= 9! × 8
APPEARS IN
RELATED QUESTIONS
Evaluate: 8!
Evaluate: 10! – 6!
Compute: `(12/6)!`
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 = 12, r = 12
Evaluate : `("n"!)/("r"!("n" - "r")!)` for n = 15, r = 8
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 (n + 3)! = 110 × (n + 1)!
Find n, if: `((17 - "n")!)/((14 - "n")!)` = 5!
Find n, if: `((15 - "n")!)/((13 - "n")!)` = 12
Find n, if: `("n"!)/(3!("n" - 3)!) : ("n"!)/(5!("n" - 5)!)` = 5 : 3
Find n, if: `("n"!)/(3!("n" - 3)!) : ("n"!)/(5!("n" - 7)!)` = 1 : 6
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)!)`
Simplify `((2"n" + 2)!)/((2"n")!)`
Simplify n[n! + (n – 1)!] + n2(n – 1)! + (n + 1)!
Simplify `("n" + 2)/("n"!) - (3"n" + 1)/(("n" + 1)!)`
Simplify `1/(("n" - 1)!) + (1 - "n")/(("n" + 1)!)`
Simplify `1/("n"!) - 3/(("n" + 1)!) - ("n"^2 - 4)/(("n" + 2)!)`
Simplify `("n"^2 - 9)/(("n" + 3)!) + 6/(("n" + 2)!) - 1/(("n" + 1)!)`
3. 9. 15. 21 ...... upto 50 factors is equal to ______.
Let Tn denote the number of triangles which can be formed using the vertices of a regular polygon of n sides. If Tn + 1 – Tn = 21, then n is equal to ______.
