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How Many 5-Digit Numbers are there? There are 90,000 five-digit numbers including the smallest five-digit number which is 10,000 and the largest five-digit number which is 99,999.∴ There are 900,000 6-digit numbers in all.There are 9000 4-digit numbers in all.
Contents
How many 6 digit numbers are there in all?
∴ There are 900,000 6-digit numbers in all.
How many 4 digit numbers are there in all?
There are 9000 4-digit numbers in all.
How many 7 digit numbers are there in all?
Find the number of seven-digit numbers which can be formed with the sum of the digits being even. Hence there are 9000000 seven-digit numbers that exist. So total number of 7 digit numbers=9999999 – 1000000+1).
How many 8 digit numbers are there in all?
Total 8-digit numbers =9,00,00,000.
How many 5 digit numbers are there with no digit repeated?
Step-by-step explanation: 9×9×8×7×6 this 5 digits are not repeated which is equal to 27216.
How many 3 digit numbers are there in all?
As we already discussed, the three digits numbers are from 100 (hundred) to 999 (nine hundred ninety-nine). Before we see the list of three digits numbers, first let us discuss how many three digits numbers we have from 100 to 999. Hence, there are 900 three-digit numbers in total.
How many 2 digit numbers are there in all?
The smallest 2-digit number is 10 and the greatest 2-digit number is 99. There are ninety, 2-digit numbers in all.
How many 6 digit numbers are there in all Brainly?
Answer. Answer: 999999 is the biggest 6-digit number and 100000 is the smallest. There are 900,000 6-digit numbers in all.
How many combinations are there for a 6 digit PIN?
Six-digit PINs not more secure than four-digit ones
A four-digit PIN can be used to create 10,000 different combinations, while a six-digit PIN can be used to create one million.
How many 6 digits numbers are there with no digit repeated?
Hence the number of six digit numbers is 1440.
Numbers Up To 5 Digits – Definition, Place Value, Expanded Form, Examples | 5 digit numbers
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What are Numbers up to 5-Digits
Smallest 5 Digit Number
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Commas in 5-Digit Numbers
How Many 5-Digit Numbers Can Be Formed
Examples on 5 Digit Numbers
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Q7How many 5digit numbers are there in all…
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- Summary of article content: Articles about Attention Required! | Cloudflare Largest 5-digit number = 99999. Largest 4-digit number = 9999. Required number = 99999−9999. = 90,000. Hence , 90,000 numbers are there in all ; Largest 5-digit … …
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How many 5-digit numbers are there in all?
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How Many 5-digit Numbers Are There in All? – Mathematics | Shaalaa.com
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How many 5 digit numbers are there that are the same class 9 maths CBSE
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combinatorics – How many $5$-digit numbers are there with digits $abcde$ such that $a < b < c \leq d \leq e$? - Mathematics Stack Exchange
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hello students welcome to Lido Learning’s question-and-answer videos my name is palibi and I teach maths and science at leto Let’s have a look at this very interesting question about mirrors so name the kind of mirror used to obtain a real and enlarge the image now before we go ahead with all the options given over here Let’s try to understand what we mean with a real image so a real image means where the rays they are the meetings actually That is, if they meet, then it’s a real the image then what is a virtual image the rays they don’t meet so in case of virtual image rays don’t meet actually so they are just extended virtually they meet virtually now let’s have a look at the question here so let’s go by the first one, a real and enlarged an image so let’s have a look at this part over here so a real and enlarged image is formed over here as you can see the rays after reflection from the mirror are actually meeting over here so it’s a real image and as you can see this is the object and this is the image so it’s enlarged as well so what type of mirror is used here it is a concave mirror answer to this is concave mirror let’s have a look at the next one a virtual and enlarged image now let’s have a look at the second picture b look at this the object is between the focus and the pole of the the concave mirror again it’s a concave mirror but as we can see the rays are getting reflected but they don’t meet exactly here right but if we virtually extend them behind the mirror then they meet here. so the image formed over here is virtual It’s erect and it’s magnified or enlarged Therefore, the answer to this one is virtual and enlarged is the concave mirror again let’s have a look on the third one a real and diminished image now has a look at this picture so we have picture c so object over here now again it is standing in front of the mirror beyond the center of curvature of the mirror where do we see is the image formed the images are formed over here where the rays are reflecting and they’re actually meeting over here. So the image form is real and it is also a diminished image as you can see the object is much bigger in size rather than the image so the image is diminished what type of mirror is used here again concave therefore the answer to the third one will also be concave mirror let’s have a look at the fourth one now so a virtual and a diminished image now this case the image the b case if you see image was virtual but it was enlarged this time we need a diminished image so Let’s have a mirror. an object in front of it as you can see the ray is getting diverged and this one is passing through the center of curvature where do the rays meet that they don’t meet so we extend them backward and they virtually meet at so they are virtually meeting in a dash therefore this is the image formed now look at the object how big it is and look at the image how small it is. Therefore, it’s a virtual image, and it is a diminished image so a virtual and diminished image is formed in this case which is d now what type of mirror this is it’s a convex mirror this time because it’s bulging out and it’s a a diverging mirror so the answer to this is convex mirror I hope you understood this if you have any further questions, please post your comments below Also, if you like this video, please don’t Forgot to subscribe to Lido channel for more such videos Thank you.
How many 5 digit numbers are there that are the same class 9 maths CBSE
Hint: By using the example given in the question, think about the number in general sense. Think about how the digits should be placed so that the number does not change after reversing its order. Then think about the possibilities of choosing any number in place of each digit.
Complete step-by-step answer :
We have to find the number of 5 digit numbers that are the same when the order of their digits is reversed.
Let us say that the five-digit number will look like the following.
$ abcba $
This number must look like above because the first digit must be the same as that of the unit place digit. The second digit must be the same as that of the ten’s place digit. Then only, they would have the same position after reversing their order.
Since, the third digit, $ c $ , is in the middle. Its position will not change by changing the order. Therefore, we do not need to consider any possibilities for it.
Now, let us consider the possibilities for each digit from the left hand side.
Digit one, i.e. $ a $ cannot be zero. Therefore, there are nine possibilities to allocate a digit to $ a $ .
Since, other digits can be zero, there are 10 possibilities to allocate a digit to both $ b $ and $ c $ , respectively.
Fourth and fifth digit, i.e. $ b $ and $ a $ will be the same as that of the first and the second digit. Therefore, there is only one possibility of allocating any digits to them.
Hence, the total possibilities of choosing a five-digit number such that it will not change after reversing its order will be the product of all the possibilities of selecting all the digits. i.e.
$\Rightarrow 9 \times 10 \times 10 \times 1 \times 1 = 900 $
Therefore, number of possibilities of selecting a five digit number such that the number will not change after reversing its order is $ 900 $
So, the correct answer is “ $ 900 $ ”.
Note: Key point in this question is to understand that $ a $ cannot be zero and hence it would have only nine possibilities. Because if $ a = 0 $ , then the given number will be a four digit number. Also, you need to understand that there is only one possibility to choose the units place and ten’s place digit as they must be the same as that of the first and the second number from the left.
How many $5$-digit numbers are there with digits $abcde$ such that $a < b < c \leq d \leq e$?
$\begingroup$
The first thing to note is that if we select k distinct digits from n digits, the digits can be arranged in a strictly ascending order in only one way
We can break up the number string into two parts
(a): $a,b,c:\;$ The ending digit($c$) can range from $3-9$ and it can be seen that if the ending digit $c$ is $n$, say, from the lower digits, there can be $\binom{n-1}2$ selections in strictly ascending order
(b): Suppose c was $3$, then we have $10-3=7$ digits for the remaining slot $e$ . These slots can either have distinct digits $\binom72$, or identical digits, $7$
Combining the two parts, the formula comes out to
$$\sum_{n=3}^{10}\binom{n-1}2\left[\binom{10-n}2+(10-n)\right] = 462$$
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