Question

How many numbers less than 10,000 be formed by using different digits from 0 to 9 repetition not allowed which are such that they are:
(a) divisible by 5
(b) divisible by 25,
What will be the corresponding answers for (a) and (b) if repetition is allowed?

Answer 1

Repetition not allowed and + 5
Numbers less than 10,000 will at the most contain 4 digits (how many digits not given) Hence any number of digits but $\ge$ 4 digits from 0 to 9 may be used. Hence the number may be of one digit or two digits or three digits and at the most of 4 digits. Other condition is divisible by 5 for (a) and divisible by 25 for (b).
(i) One digit number (5) = 1
(ii) Two digit numbers = ||5+||0
i.e. those having 5 in the unit place or 0 in the unit place = ${(}^9{P}_1{-}^8{P}_0){+}^9{P}_1=(9-1)+9=17$
(iii) Three digit numbers =||5+||0 as above
=${(}^9{P}_2{-}^8{P}_1){+}^9{P}_2=2(9.8)-8=136$
(iv) Four digit numbers = ||5+ ||0 as above =${(}^9{P}_3{-}^8{P}_2){+}^9{P}_3=2(\mathrm{9.8.7})-8.7$
= $\left(8.7\right).17=56\times 17=952.$
Hence total is $1+17+136+952=1106$.
Repetition allowed and + 5
(i) 1 digit 5 1 only
(ii) 2 digit $\ne 0|9|5+|9|0=9+9=18$
After fixing 5 in the unit place the first place can be filled again by 9 ways (excluding zero) as 5 can again come because repetition is allowed.
(iii) 3 digit $\ne 0|9\times 10|5+|9\times 10|0=180$
(iv) 4 digit $\ne 0|9\times 10\times 10|5+|9\times 10\times 10|0=1800$
Total $=1+18+180+1800=1999.$
(b) Numbers divisible by 25.
Repetition not allowed and +25
1 digit No 0
2 digit 25 or 50,75 3
3 digit $\ne 0|7|25and|7|75and|8|50$ $7+7+8=22$.
4 digit $\ne 0|7\times 7|25and\ne 0|7\times 7|75and|8\times 7|50$ $49+49+56=154$
Two digits have been fixed in R.H.S., 0 cannot come in L.H.S. and hence we are left with only 7 choices for first number in L.H.S. For 2nd number from L.H.S. since 0 can come in between we have again 7 choices.
Total = 3+22+154=179
Repetition allowed and +25
1 digit 0
2 digit 25, 50, 75 3
3 digit $9\times 4=36$
4 digit $4\times 90=360$
Total =$3+36+360=399.$