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2. You’re 100% correct. There are 10 possible numbers for the first digit, and then you can’t use that number again, so 9 for the second, and using the same logic, 8 for the third and 7 for the fourth. That means there’s 10 × 9 × 8 × 7 = 5040 10 × 9 × 8 × 7 = 5040 combinations. Divide this by the number of ways to order each one, 24 ...
24" combinations" >"the possible combinations are" "using the 4 digits 1234" ((1,2,3,4),(1,2,4,3),(1,3,2,4),(1,3,2,4),(1,3,4,2),(1,4,2,3),(1,4,3,2))=6((2,1,3,4),(2,1 ...
The third choice comes from 8 possibilities and the fourth from 7 possibilities. Now we multiply these together: 10 x 9 x 8 x 7 = 90 x 56 = 5040. That's the number of permutations. No digits repeat, but 0123 is different from 0321.
Now if you add the hex letters a,b,c,d,e,f then now there are 16 possible combinations for each digit (the 10 numbers + 6 letters). So I think it stands to reason that you take the number of possible combinations for a digit, 16, and raise it by the number of digits in your code. So 16^8 for an 8 digit code, 16^10 for a 10 digit code, 16^16 for ...
In that case, the number of 4 -digit combinations is given by. 10 ⋅ 9 ⋅ 8 ⋅ 7 = 5,040. Please read the explanation, because the answer is either 10,000 or 5,040. Since there are 10 choices for each digit, the number of 4- digit combinations is given by 10*10*10*10 = 10,000, UNLESS using a digit means it cannot be used again. In that case ...
The total number of combinations is (104) (10 4). For each combination there are 4! 4! different arrangements. Exactly 1 1 of these arrangements is strictly increasing. Exactly 1 1 of these arrangements is strictly decreasing. Hence the number of valid arrangements is (104) ⋅ (4! − 1 − 1) (10 4) ⋅ (4! − 1 − 1).
How many 4-digit numbers ($0000-9999$; including $0000$ and $9999$) can be formed in which the sum of first two digits is equal to the sum of last two digits? Assumption : every number is valid even if it starts with a zero. For ex: $1230, 0211, 4233$ and so on...
We form all possible $4$-digit numbers using $1,2,3,4,5$. How many such numbers are possible? What is the sum of all these 4-digit numbers? How do the answers change if we use $1,1,2,3,4$ instead. My try: $5$ choices for each position. Therefore $5\cdot5\cdot5\cdot5= 625$ $4$ choices as $1$ is repeated $2$ times. Therefore $4\cdot4\cdot4\cdot4 ...
For $(1)$ the number of such numbers is $2 \cdot (4!/3!)=8$ (you have calculated) For $(2)$ the number of such numbers is $4!/(2! \cdot 2!) =6$ So adding $(1)$ and $(2)$ we get the total number of such numbers i.e $8+6=14$ So the probability is $14/2^4=0.875$. So you should be very careful how we can fill those empty places after placing the ...
To find the probably, all the favourable outcomes (the number of 4-digit number combinations that start with 1 or 2, that have at least 3 of the same digit), must be divided by the total amount of 4-digit number combinations. The range of 4-digit numbers is between 1000 and 9999, as a 4-digit number cannot start with 0. (0999 cannot occur, 0023 ...