How many six digit positive integers that are divisible by 3 can be formed using the digits 0,1,2,3,4,5 and 6 without any of the digits getting repeating?

Asked by Last Modified  

23 Answers

+2 Mathematics Permutations

Follow 2
Answer

Please enter your answer

Sujoy Das

JU-IITKGP & 10+ years experienced for boards, JEE,NEET,OLYMPIAD & 5-12

216
Comments
Dislike Bookmark

Tutor

216
Comments
Dislike Bookmark

B.tech

Last two digits should be 30,03,12,21,36,63,45,54 to divisible by three... so total possible arrangements could be = 4x(4x3x2x1)x2 = 192 digits
Comments
Dislike Bookmark

Tutor

192
Comments
Dislike Bookmark

Experienced Mathematics com ECE Educator

Answer is 1320 Explaination: To form 6 digit number which is divisilble by 3. first condition is sum of number is divisible by 3i.e., 0+1+2+3+4+5+6=21 which is divisible by 3. since out of 7 digit , only 6 digit has to be placed, one digit has to be removed. (which should be divisible by 3) So, by...
read more
Answer is 1320 Explaination: To form 6 digit number which is divisilble by 3. first condition is sum of number is divisible by 3i.e., 0+1+2+3+4+5+6=21 which is divisible by 3. since out of 7 digit , only 6 digit has to be placed, one digit has to be removed. (which should be divisible by 3) So, by removing 0 and 3 from series the total number will always be divisible by 3. 1.)) by removing 0 from series(1,2,3,4,5,6):-we get 6 digits can be arranged in 6! times = 720 2.)) by removing 3 from series(0,1,2,4,5,6):- 0 cannot take first place from left so without it, we have remaining 5 number and now including 0 in other 5 places we can arrange 5 digits in 5! ways. = 5*5!= 5*120=600 3)) In total 720+600=1320 read less
Comments
Dislike Bookmark

Tutor

here i have posted the answer.
Comments
Dislike Bookmark

1. If '0' is not used then it can be arranged in 6 possible ways 6!=720 numbers 2.if we don't use '3' and '0' cannot be the first digit, hence we have only 5 digits as 1st number they are 1,2,4,5,6 Then remaining 5 digits can be arranged in other 5 places hence we will have 5 digits in 5 ways 5*5!=600...
read more
1. If '0' is not used then it can be arranged in 6 possible ways 6!=720 numbers 2.if we don't use '3' and '0' cannot be the first digit, hence we have only 5 digits as 1st number they are 1,2,4,5,6 Then remaining 5 digits can be arranged in other 5 places hence we will have 5 digits in 5 ways 5*5!=600 numbers By adding 1 and 2 we will have 720+600=1320 numbers read less
Comments
Dislike Bookmark

Tuition for girls and small kids

720
Comments
Dislike Bookmark

ans is 1920 divisibility test of 3 is the sum of digits should be divisible by 3. So sum of 0,1,2,3,4,5,6 is 21. case 1: when 1,2,3,4,5,6 numbers are taken possible number will be factorial 6=120 Case 2: when 0,1,2,4,5,6 are taken possible numbers will be 5*5! (as 0 cannot come as first digit otherwise...
read more
ans is 1920 divisibility test of 3 is the sum of digits should be divisible by 3. So sum of 0,1,2,3,4,5,6 is 21. case 1: when 1,2,3,4,5,6 numbers are taken possible number will be factorial 6=120 Case 2: when 0,1,2,4,5,6 are taken possible numbers will be 5*5! (as 0 cannot come as first digit otherwise that would not be an 6 digit no. Case3 : when 0,1,2,3,4,5 numbers are taken possible number will be 5*5! Ans is 120+(2*5*5!) read less
Comments
Dislike Bookmark

C++ Tutor

ones place can be filled in 6 different ways tens place can be filled in 5 different ways hundredth place can also be filled in 5 different ways ans should be 6!*5!*5!
Comments
Dislike Bookmark

View 21 more Answers

Related Questions

In how many ways can the letters of the word "ASSASSINATION" be arranged so that all the S's are together ?
the word "ASSASSINATION" has 4S , 3A , 2I , 2N , T , O , 4S are together. this is considered as one block as 1 letter. we have 3A , 2I, 2N , 4S , T , O therefore , Number of words = Number of Permutations...
Kumar Upendra Akshay
What is the equation of tangent hyperbola?
y= mx + sqrt(a*a*m*m - b*b)
Kartika
0 0
8
There are 4 letters and 4 addressed envelopes. find the probability that all the letters are not dispatched in the right envelope.
4! (1/2! - 1/3! +1/4! ) = 9
Kumar Upendra Akshay
0 0
8
Guys Hi! I hope your preparation is going well for pre-board exam. Please post any query in math, if you have!
If you are getting ready for the exam then few things are seems very important please go through it without fail: these are from NCERT book .. Exercise 5.1 (11, 13, 15, 19, 20) Exercise 5.2 (example...
Nilaish
1 0
6
What is the value of cos 45 + sin 90?
0.7 + 1= 1.7
Asif

Now ask question in any of the 1000+ Categories, and get Answers from Tutors and Trainers on UrbanPro.com

Ask a Question

Related Lessons

Henry Fayol's 14 Principles of Management-HSC,Class 12,OC,Business Studies
14 Principles of Management by Henry Fayol

Rajiv Vadera

0 0
0
How to solve 2D, 3D problems?
Let us take a problem in 3D? Consider the triangle A(0,0,0), B(1,1,1) and C(0,0,1). What is the area of this triangle? Solution: As we cannot plot these points, we analyse them as where they lie in...

Harendran Bala

0 0
0
Global Enterprises at a Glance, MNC and other international forms of business
Global Enterprises: These companies have many markets other than their own country market (domestic market) to sell their product. Their products are available nearly all over the world. Why has globalization...

Educator Titir S.

0 0
0
Solution to a integration problem
∫ x^4 / ( x² + 1 ) dx As the degree of numerator is higher than that of the denominator, let's divide x^4 by ( x² + 1 ). = ∫ x² - 1 + 1/(x² + 1 ) dx = ∫ x² dx - ∫1•dx + 1/ ( 1 + x² ) dx =

B.sudhakar

3 0
0
Mean value theorem in Calculus
Mean value theorem in simple terms: 1. If a real valued function f(x) is continuous over an closed interval 2. And this function is differentiable over an open interval (a,b) 3. f(a) ≠ f(b) : (...

B.sudhakar

7 0
0

Recommended Articles

Meet Urmila, an MBBS tutor from Bangalore

Urmila is a passionate teacher with over 8 years of experience in teaching. She is currently pursuing her Ph. D. She provides classes for Class 11, Class 12, MBBS and Medical tuition.  Urmila began her career in teaching long before she became a teacher. She used to provide classes for foreign national students in her college...

Read full article >

Meet Radhe Shyam Burman, an MBA Tutor from...

Radhe Shyam is a highly skilled accounts and finance trainer with 8 years of experience in teaching. Accounting is challenging for many students and that’s where Radhe Shyam’s expertise comes into play. He helps his students not only in understanding the subject but also advises them on how to overcome the fear of accounts...

Read full article >

Meet Mohammad Wazid, a skilled trainer for...

Mohammad Wazid is a certified professional tutor for class 11 students. He has 6 years of teaching experience which he couples with an energetic attitude and a vision of making any subject easy for the students. Over the years he has developed skills with a capability of understanding the requirements of the students. This...

Read full article >

Meet Swati, a Hindi Tutor from Bangalore

Swati is a renowned Hindi tutor with 7 years of experience in teaching. She conducts classes for various students ranging from class 6- class 12 and also BA students. Having pursued her education at Madras University where she did her Masters in Hindi, Swati knows her way around students. She believes that each student...

Read full article >

Find Class 12 Tuition near you
Class 12 Tuition in Delhi
Class 12 Tuition in Bangalore
Class 12 Tuition in Hyderabad
Class 12 Tuition in Chennai
Class 12 Tuition in Kolkata
Class 12 Tuition in Mumbai
Class 12 Tuition in Pune
Class 12 Tuition in Lucknow
Class 12 Tuition in Noida
Class 12 Tuition in Gurgaon
Online Tuition Class 12

Looking for Class 12 Tuition ?

Learn from the Best Tutors on UrbanPro

Are you a Tutor or Training Institute?

Join UrbanPro Today to find students near you