Question

A number of $4$ different digits is formed by using the digits $1,2,3,4,5,6,7$ in all possible ways. Find
(a) How many such numbers can be formed?
(b) How many of them are greater than $3400$?
(c) How many of them are exactly divisible by $2$?
(d) How many of these are exactly divisible by $25$?
(e) How many of these are exactly divisible by $4$?

Answer 1

1, 2, 3, 4, 5, 6, 7 : 7 digits in all and we have to use 4 digits
(a) ${}^7{P}_4=\frac{7!}{3!}=7\times 6\times 5\times 4=840$.
(b) Numbers greater than 3400 will have, 4 or 5 or 6 or 7 in the first place i.e. there are 4 ways of filling the first place. (i.e. 4 ways) Having filled the first place say by 4 we have to choose 3 digits out of the remaining 6 and the number will be
${}^6{P}_3=\frac{6!}{3!}=6\times 5\times 4=120$
Therefore total of such numbers by fundamental theorem will be
$4\times 120=480$. ...(1)
Numbers greater than 3400 can be those which have 34, 35, 36, 37 in the first two places (i.e. 4 ways).
Having filled up 34 in the first two places we will have to choose 2 more out of remaining 5 and the number will be
${}^5{P}_2=\frac{5!}{3!}=5\times 4=20$
Therefore total as above will be
$20\times 4=80$. ...(2)
Hence all the numbers greater than 3400 will be
480 + 80 = 560, by (1) and (2).
(c) The numbers will be divisible by 2 if the last digit is divisible by 2 which can be done in 3 ways by fixing 2 or 4 or 6 and the remaining 3 places can be filled up out of remaining 6 digits in ${}^6{P}_3$ ways.
Hence the required no.
$3\times x^6{P}_3=3\times 120=360$.
(d) A number will be divisible by 25 if the last two digits are divisible by 25 and this can be done in two ways for either 25 or 75 can be there and remaining two places out of 5 digits can be filled in ${}^5{P}_2$ ways.
Hence the required number
$=2{\times }^5{P}_2=2\times 20=40$.
(e) A number is divisible by 4 if the last two digits are divisible by 4 which can be done in 10, ways (12, 16, 24, 32, 36, 52, 56, 64, 72, 76).
Hence number = $10{\times }^5{P}_2=10\times 20=200$.