Question

A number of 6 different digits is formed by using the digits 0, 1, 2, 3, 4, 5. Find
(a) How may such numbers can be formed?
(b) How many of these are even?
(c) How many of these are divisible by 4?
(d) How many of these are divisible by 25?

Answer 1

(ii) Care has to be taken because of 0 as 032451 is not of six digit but of five digit.
(a) $6!-5!=720-120=600$
5! corresponds to the number which start with zero.
(b) Even numbers. These numbers will have $0,2,4$. in the end.
Having 0 in the end we have to arrange the remaining 5 in 5! = 120 ways.
Having 2 in the end. 5! - 4! (1st 0 ............ last 2)
= 120 - 24 =96
Having 4 in the end = 96 as above (1st 0 ..... last 2)
Total = $120+96+96=312$
(c) Divisible by 4
The last two digits will be divisible by 4. They will be of the type (12, 24, 52) or (04, 20, 40) we have to arrange remaining 4 digits.
For 1st bracket the remaining 4 digits will include 0 which cannot occupy the first place.
Hence $4!-3!(0.....)=18$
Thus total =$18+18+18=54$
For each of 2nd bracket we will have to arrange remaining 4 non-zero digit which can be done in 4! = 24 ways.
Total for 2nd brackets = $24+24+24=72$
Hence their will be =$54+72=126$ numbers divisible by 4.
(d) Divisible by 25
The last two digit be will be 25 or 50. Arranging as above for 50 it will be 4! = 24 and for 50 it will be $4!-3!(0.....)=18$
Total divisible by 25 is $24+18=42$