How many different 6 digit numbers can be formed with the digits 3,5,6,7,8, 9

question kya hai how many different numbers of 6 digits can be formed from the digits 456789 when repetition of the digits is not allowed theek hai Hamen kitne digit ka number 1 2 1 2 3 4 5 6 6 digit number banana hai Hamen all repetition is allowed Nahin Hai To jo pahle usko karne ke kitne tarike 4 5 6 7 8 9 10 tarike pahle breast ko badhane ki dusri place karne ke kitne tarike bacche 5 Kyon Kyunki repetition allowed Malik latest Yahan per 5 a gaya ab yah 5 Yahan Ke Tu Yahan to are hi Nahin Sakta to kitne tarike bacche sirf Yahan per Yahan per 3 Yahan Ke Liye Do Aur Yahan Ke Liye Ek Tarika 2 final answer mera kya aayega 6 into 5 into 4 into 3 into 2

1 theek hai yah Kitna Nikal Ke Jaega mere pass 6:30 into yoga Ye Jaega aapke pass 722 yah Ho Gaya final answer to Kabhi iji aur ekdum basic question hai bus Tarah Se Garam analytical is ko Soch Le to easily

question to find out how many different 6 digit numbers can be formed using all of the following digits of a given 6th it but not not all of them are distinct so number of arrangements of N distinct objects is nothing but n factorial but not distinct will have to use a different formula which is number of arrangements of objects animal are same n to ask same

comma and 3 are same Sone after anal are saying what is meant by this is that there and one object that are similar to one another and two objects that are similar to entry of these and similar to another so this formula is actually n factorial by anyone factorial into N2 factorial of tell NM with what is mentir we can understand this by looking at these questions at this number as we can see that here 23 is a sem 3 force RC so number of arrangements

of 33 44 45 is 6 factorial first total number 6 / 2 factorial 323 is a same and 34 others and to have told him that none of the rest of them are saying we can you so this is 6 into 5 into 4 into 3 into 2 into 1 6 factorial to have nothing but 21213 factorial is 3 into 2 into 1 3 and 3A cancel 21 to get cancer one and one gets cancelled what is the centre nh21 forget cancel easement 26 into 5 into 5 into 10 into 10 so

there are 60 possible arrangements of these numbers option d is the correct answer

Answer

Verified

Hint: We are given a problem in which there are 6 different digits given such as to form a six digit number. Then in the second sub question that number should be divisible by 5 that is on the last two places can be filled by either 0 or 5. And in the third sub question we need to find the numbers that are not divisible by 5. That is, the last two places should have any digit other than 0 and 5.

Complete step by step solution:
We will solve the problem step by step.
I)First let's find the numbers that can be formed.
Now we cannot place 0 on the highest place value because that will not be the six digit number. So we have five other places for it.
Now one of the remaining 5 numbers can be placed in 5 ways. Similarly one by one the numbers will be placed. So the total numbers will be,
\[5 \times 5!\]
That is \[5 \times 120 = 600\]
600 different numbers will be there.
II) They may have 5 in the last place and as above we will have 5! = 120 ways. These will also include numbers which will have zero in the first place. Therefore the numbers having zero in 1st and 5 in last place will be 4!
Therefore 6 digit numbers having 5 in the end will be
$5! -4! = 120 - 2.4 = 96$.
Therefore the total number of 6 digit numbers divisible by 5 is
120 + 96 = 216.
Only 216 numbers are there; those are divisible by 5.
III) At last the numbers that are not divisible by 5 can be found directly by subtracting the numbers that are divisible from total numbers.
So that is \[600 - 216 = 384\]
Thus we completed the solution.

Note: Note that placing zero on the highest place value will not fulfill the condition of the 6 digit number mentioned. And in order to find the number not divisible by 5. we will not shuffle again. we will just remove the numbers that are divisible.
The other divisible numbers can also be found just by using the divisibility tests of the respective numbers requirement.

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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