How many numbers can be formed from digits 1 2 3 9 if repetition of digits is not allowed?
|
Show
How many numbers can be formed from digits 1, 3, 5, 9 if repetition of digits is not allowed?Answer Verified
Hint: The numbers can be 1 digit, 2 digits, 3 digits, and 4 digits as well. The number of 1 digit numbers that can be formed by using the digits 1, 3, 5, and 9 are 4. For a 2 digit number, as we have 4 intakes and 3 intakes at 1st place and 2nd place respectively. So, the number of 2 digit numbers is 12. For 3 digit numbers, as we have 4 intakes, 3 intakes, and 2 intakes at 1st place, 2nd place, and 3rd place respectively. So, the number of 3 digit numbers is 24. For 4 digit numbers, as we have 4 intakes, 3 intakes, 2 intakes, and 1 intake at 1st place, 2nd place, 3rd, and 4th place respectively. So, the number of 4 digit numbers is 24. Now, the total numbers that can be formed are the summation of 1 digit numbers, 2 digit numbers, 3 digit numbers, and 4 digit numbers. Complete step-by-step answer: Here we have to find the number of numbers that can be formed from digits 1, 3, 5, and 9. The numbers can have 1 digit, 2 digits, 3 digits, and 4 digits as well. So, we have to find the numbers step by step. Let us continue with 1 digit number. As we have 4 digits, so the number of single-digit numbers is 4. Now, we have to find the number of 2 digit numbers. For the first place, we can take any of the four digits given. But we cannot take any of the 4 digits given in the second place because repetition is not allowed here. So, we have 3 intakes for second place. Let us understand with a diagram. The total number of 2 digit numbers \[=4\times 3=12\] . Now, we have to find the number of 3 digit numbers. For the first place, we can take any of the four digits given. But we cannot take any of the 4 digits given in the second place because repetition is not allowed here. So, we have 3 intakes for second place. Similarly, for the 3rd place, we have to choose any one digit of the remaining two digits. Let us understand with a diagram.  The total number of 3 digit numbers \[=4\times 3\times 2=24\] . Now, we have to find the number of 4 digit numbers. For the first place, we can take any of the four digits given. But we cannot take any of the 4 digits given in the second place because repetition is not allowed here. So, we have 3 intakes for second place. Similarly, for the 3rd place, we have to choose any one digit of the remaining two digits and proceed with 1 intake for the 4th place. Let us understand with a diagram. The total number of 4 digit numbers \[=4\times 3\times 2\times 1=24\] . Total numbers formed = 1 digit numbers + 2 digit numbers + 3 digit numbers =4+12+24+24 =64 Hence, 64 numbers can be formed. Note:This question can also be solved using a combination method. We also know that the rearrangement of n digits is \[n!\] . To find the number of 2 digit numbers, we have to take any two of the four digits given and also rearrangement is possible. That is the number of 2 digit numbers are \[^{4}{{C}_{2}}\times 2!=12\] . Similarly, for 3 digit numbers, we can take any three of the four digits, and also rearrangement is possible. That is the number of 3 digit numbers are \[^{4}{{C}_{3}}\times 3!=24\] . Similarly, for 3 digit numbers, we can take any four of the four digits, and also rearrangement is possible. That is the number of 3 digit numbers are \[^{4}{{C}_{4}}\times 4!=24\] . The digits chosen must sum to a multiple of 3, but not to a multiple of 9. If no repeated digits are allowed, the combinations of digits that have the appropriate sums are Nội dung chính
If digits are allowed to be repeated, there are 28 choices. When digits are repeated, the number of possible variations in the digit
sequence is reduced. The choices are Altogether, there are 295 different numbers that can be made with these sets of digits. Right 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 AP44Explanation for the correct option:To find the total numbers:Given digits1, 2,3,4 and repetition is not allowed.Ways to choose the first digits =4Ways to choose the second digits=3Ways to choose the third digits =2Ways to choose the fourth digits =1 (adsbygoogle = window.adsbygoogle || []).push({}); Thus total numbers =4×3×2×1=P44Hence the correct option is A.TextbooksQuestion PapersHomequestion says that how many numbers can be formed from the digit 1 2 3 and 9 if repetition of digits is not allowed so the digits given to us is 123 and night superstar ful the maximum number which can be formed from this four digits is 9321 this can be the maximum number can be found from this four digit number we cannot form greatest number that this number with this 4 digit numbers so let us see first the 4 digit numbers that can be formed from this number so 4 digit number which can be formed from this digits are there for places and repetition is not allowed so for place we have four options to fill this place that is either one can come to can come 3 can come on come addition of digit is not allowed if we have placed a digit hair soap for this please only three digits are left similarly for this two students are left and for the last digit only one digit is left so it is and what are the fundamental principle of multiplication will be multiplied so that is 4 into 3 into 2 into 1 is 24 ways are there so it is 2444 digit numbers can be formed from this four digits now that is 3 how many 3 digit numbers can be formed from the digits Shri places and in the first place therefore option to fill this first place and similarly there are three options for failed to fill this place and there are two option to fill this place that is also 24 number so there are 24 numbers of 3 digits can be possible from these four digits given to us that AC two digit number now Chhotu digits number so there are two places in two digit number for this place four options are available and for this place there are three options are available so it is 12 numbers so well numbers of two digits can be possible with this four digits and one digit number will be the only one possible way that is 4 and which can be either 123 online this phone number one digit is possible it is phone numbers so total numbers that is possible with given four digits are 24 + 24 + 12 + 4 that is equal to 25 + 2448 4850 1664 number 7 64 numbers are possible with the given 1 2 3 and 9 digits everybody son of digit is not allowed How many three digit numbers can be formed using the digits 1 2 3 if repetition of digit is allowed?Answer: 125 As repetition is allowed, So the number of digits available for B and C will also be 5 (each). How many 3 digit numbers can be formed if repetition is not allowed?There are 504 different 3-digit numbers which can be formed from numbers 1, 2, 3, 4, 5, 6, 7, 8, 9 if no repetition is allowed. Note: We can also use the multiplication principle to answer this question. How many 3 digit numbers can be formed using 12345 repetition?∴ Total number of 3-digit numbers = 3×4×5=60. How many numbers can be formed from the digits 2 4 6 9 if repetition of digits is not allowed an 64?1 Answer. ∴ Total numbers = 4 + 12 + 24 + 24 = 64 numbers. How many combinations can you make with the numbers 1 2 3 with repetition?That is a total of 7 combinations.
How many numbers can be formed using the digits 1 2 5 7 9 if repetition of digits is not permitted?Hence, 64 numbers can be formed.
How many 3Answer: 24
As repetition is not allowed, So the number of digits available for B = 3 (As one digit has already been chosen at A), Similarly, the number of digits available for C = 2.
How many ways the three digits 1/2 and 3 can be arranged if repetition is allowed?So I take some particular numbers, like 1,2,3 and say that, well, 1 can go in 3 places, 2 in 2 places and 3 in 1 place, so by multiplication principle, there are 6 ways of forming a 3-digit number with 1,2,3. But there are 4 different numbers. So the number of 3-number combinations are- (1,2,3),(1,2,4),(1,3,4),(2,3,4).
|
