Advertisements
Advertisements
Question
Evaluate: `("n"!)/("r"!("n" - "r"!)` For n = 8, r = 6
Advertisements
Solution
n = 8, r = 6
`("n"!)/("r"!("n" - "r"!)) = (8!)/(6!(8 - 6!))`
= `(8 xx 7 xx 6!)/(2!6!)`
= `(8 xx 7)/(2!)`
= `(8xx7)/(1xx2)`
= 28
APPEARS IN
RELATED QUESTIONS
A teacher wants to select the class monitor in a class of 30 boys and 20 girls, in how many ways can the monitor be selected if the monitor must be a boy? What is the answer if the monitor must be a girl?
How many three-digit numbers can be formed using the digits 2, 3, 4, 5, 6 if digits can be repeated?
If numbers are formed using digits 2, 3, 4, 5, 6 without repetition, how many of them will exceed 400?
Compute: `(12!)/(6!)`
Compute: 3! × 2!
Compute: `(6! - 4!)/(4!)`
Write in terms of factorial:
5 × 6 × 7 × 8 × 9 × 10
Write in terms of factorial:
3 × 6 × 9 × 12 × 15
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)! = 110 × (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
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: `(9!)/(3!6!) + (9!)/(4!5!) = (10!)/(4!6!)`
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.
