Advertisements
Advertisements
Question
If numbers are formed using digits 2, 3, 4, 5, 6 without repetition, how many of them will exceed 400?
Advertisements
Solution
Case I: Three-digit numbers with 4 occurring in hundred’s place:
100’s place digit can be selected in 1 way.
Ten’s place can be filled by anyone of the number 2, 3, 5, 6.
∴ 10’s place digit can be selected in 4 ways.
Unit’s place digit can be selected in 3 ways.
∴ Total number of numbers which have 4 in 100’s place = 1 × 4 × 3 = 12
Case II: Three digit numbers more than 500
100’s place digit can be selected in 2 ways.
10’s place digit can be selected in 4 ways.
Unit’s place digit can be selected in 3 ways.
∴ Total number of three digit numbers more than 500 = 2 × 4 × 3 = 24
Case III: Number of four-digit numbers formed from 2, 3, 4, 5, 6
Since, repetition of digits is not allowed
∴ Total four digit numbers formed = 5 × 4 × 3 × 2= 120
Case IV: Number of five digit numbers formed from 2, 3, 4, 5, 6
Since, repetition of digits is not allowed
∴ Total five digit numbers formed = 5 × 4 × 3 × 2 × 1 = 120
∴ Total number of numbers that exceed 400 = 12 + 24 + 120 + 120 = 276
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 he select a student if the monitor can be a boy or a girl?
Evaluate: 8! – 6!
Evaluate: (8 – 6)!
Compute: `(12/6)!`
Compute: 3! × 2!
Compute: `(9!)/(3! 6!)`
Write in terms of factorial:
3 × 6 × 9 × 12 × 15
Write in terms of factorial:
6 × 7 × 8 × 9
Evaluate: `("n"!)/("r"!("n" - "r"!)` For n = 8, r = 6
Find n, if `"n"/(6!) = 4/(8!) + 3/(6!)`
Find n, if: `((17 - "n")!)/((14 - "n")!)` = 5!
Find n, if: `((15 - "n")!)/((13 - "n")!)` = 12
Show that: `(9!)/(3!6!) + (9!)/(4!5!) = (10!)/(4!6!)`
Find the value of: `(5(26!) + (27!))/(4(27!) - 8(26!)`
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?
