Find the number of 6-digit numbers using the digits 3, 4, 5, 6, 7, 8 without repetition. How many of these numbers are divisible by 5
Solution
There are 6 different digits and we have to form 6-digit numbers, i.e., n = 6, r = 6
If no digit is repeated, the total numbers with 6-digits can be formed = nPr
= 6P6
= 6!
= 6 x 5 x 4 x 3 x 2 x 1
= 720.
Since the number is divisible by 5, the unit's place of 6-digits number can be filled in only one way by the digit 5.
Remaining 5 positions can be filled from the remaining 5 digits in 5P5 ways.
Hence, the total number of 6-digit numbers divisible by 5 = 1 x 5P5 = 5!
= 5 x 4 x 3 x 2 x 1
= 120.
Concept: Permutations - Permutations When Repetitions Are Allowed
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Solution
The correct option is A
180000
Next four places can be filled with any of the 10 digits.
After filling the first five places, the last place can be filled in following ways.
CASE I : If the sum of first 5 digits is of 5k type, then rightmost digit would be either 0or5.so, units place can be filled in 2 ways
CASE II : If the sum of first 5 digits is of 5k+1 type, then rightmost digit would be either 4or9 .so, units place can be filled in 2 ways
CASE III : If the sum of first 5 digits is of 5k+2 type, then rightmost digit would be either 3or8.so, units place can be filled in 2 ways
CASE IV : If the sum of first 5 digits is of 5k+3 type, then rightmost digit would be either 2or7.so, units place can be filled in 2 ways
CASE V: If the sum of first 5 digits is of 5k+4 type, then rightmost digit would be either 1or6.so, units place can be filled in 2 ways
∴total such numbers possible =9×104×2=180000