How many 5 digit numbers exist such that sum of digits is equal to 9?
So, there are 495 numbers exist whose sum of digits is 9 for a 5 digit number.
How many counting numbers have four distinct nonzero digits such that the sum of the four digits is?
The number of possible arrangements (permutations) of 5789 = 4*3*2*1 = 4! = 24. Therefore, there are 24 numbers formed from the set of nonzero digits such that 4 distinct digits have a sum of 29.
How many 5 digit numbers are there such that sum of their digits is even?
Yes there are 45000 five digit numbers whose sum of the digits will be a even number.
How many distinct numbers are there with 5 digits?
Therefore, there are 90,000 unique 5-digit numbers possible.
How many strings of 5 digits have the property that the sum of their digits is 7?
How many strings of 5 digits have the property that the sum of their digits is 7? total = 10 + 20 + 5 + 60 + 30 + 30 + 60 + 20 + 20 + 30 + 20 + 20 + 5 = 330. So, option (B) is correct. Total number of combinations are (n-1+r)C(r) = (5-1+7)C7 = 11C7 = 11C4 = 330.
How many four digit number that can be formed with four non zero digit?
The first digits can be any one of 4, 7, or 9 and the other three digits can be any of the four. This means that there are: 3x4x4x4 = 192 possible four digit numbers.
How many 5 digit numbers are there in which odd positions are filled with even digits?
49999 – 5000 + 1 = 45,000. There are 45000 even five digit numbers. There are 5 odd digits, so there are 5 ways to choose each digit, and the number of five digit numbers composed of only odd digits is 5^5 = 3125.
How many 5 digit numbers with distinct digits are there such that in each number the digits are in descending order from left to right?
The resulting 5 digit numbers are all possible 5-digit numbers in descending order. Therefore, there are [10!/(5! 5!)] = 252 5-digit numbers in strictly descending order.
How many 5 digit numbers are there with distinct digits can be formed using the digits 1 2 5 5 4?
How many 5 digit even numbers with distinct digits can be formed using the digits 1, 2, 5, 5, 4? How many 5 digit even numbers with distinct digits can be formed using the digits 1, 2, 5, 5, 4? Explanation: The 5 digit even numbers can be formed out of 1, 2, 5, 5, 4 by using either 2 or 4 in the unit’s place.
How many strings of 5 digits have the property that the sum?
How many strings of 5 digits have the property that the sum of their digits is 7? total = 10 + 20 + 5 + 60 + 30 + 30 + 60 + 20 + 20 + 30 + 20 + 20 + 5 = 330.
How many members of a shall be divisible by 3 or by 5 or by both 3 and 5?
999} which are divisible by 3 are 999 / 3 (A)= 333 . From set A numbers {5,10,…… 995,1000} which are divisible by 5 are 1000 / 5 (B)= 200. From set A numbers {15, 30.990} which are divisible by 3 and 5 are 990 / 3 * 5 (A ∧ B)= 990 / 15 = 66….Discuss it.
| A | x8 – 71 and x8 – 71 |
|---|---|
| D | x8 + 73 and x8 + 73 |
How many 3 digit numbers have 4 distinct non zero digits?
The only possible counting number that has four distinct non-zero digits is 9875. This has 4! ways to arrange the digits leading to 24 cases. How many 3-digit number have distinct digits such that one digit is the average of the other two?
How many 5-digit numbers have an even sum of digits?
But, we are interested in the numbers whose sum is even so it will be half of the total five digit numbers, so the answer is 1 2 × ( 9 × 10 × 10 × 10 × 10) which is 45000. There are 90000 5 -digit numbers, exactly half have an even sum of digits (base 10 has equal numbers of even and odd digits)
How many ways can you choose the first digit of a number?
The first digit can be chosen in 9 ways (since it can’t be a 0), the second in 10 ways, the third in 10 ways, the fourth in 10 ways, and the fifth in five ways (the last digit must be one of the five even digits if the first four digits add up to an even number, and odd otherwise, so that the total sum is even).
Why do s and E partition the collection of 5 digit numbers?
S and E partition the collection of all 5 digit numbers and S does not contain 99999. Adding 1 to elements of S gives elements of E. This function is from S to E and is in fact a bijection and so both sets contain equal number of elements.