How many 6 digit numbers can be formed using the digits 1 to 6 without repetition such that the number is divisible by the digit at its units place?
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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 Show
SolutionThere 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 Is there an error in this question or solution? APPEARS INRight on! Give the BNAT exam to get a 100% scholarship for BYJUS courses No worries! We‘ve got your back. Try BYJU‘S free classes today! No worries! We‘ve got your back. Try BYJU‘S free classes today! No worries! We‘ve got your back. Try BYJU‘S free classes today! Solution The correct option is A 180000First place from left cannot be filled with 0. (adsbygoogle = window.adsbygoogle || []).push({}); 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=180000How many 6 digit numbers can be formed from 1 to 6 which are divisible by 4?Thus, 96 6-digit numbers can be formed that are divisible by 4.
How many 6 digit numbers can be formed using the digits without repetition with repetition?The digits are all non-zero: Choose 6 symbols from {1,2,3,4,5,6,7,8,9} and permute them: (96)⋅6! =60480.
How many six digit numbers can be formed using the digits 1 to 5 without repetition such that the number is divisible by the digit at its units place?So the number of 6 digit numbers divisible by the number in the units place is: 384+24+18=426. Was this answer helpful?
How many 6 digit numbers can be formed by first 6 whole numbers with repetition which are divisible by 2?Hence, a total of 19,440 6-digit numbers, divisible by 2, can be formed by the given set of numbers, with repetition allowed.
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