How many different numbers can be formed fro the set of seven numbers (1,2,3,4,5,6,7) if
a) repitition is allowed
b) repitition is not allowed
c) repitition is allowed and the tird digit must be different from the second digit?
a) digits can repeat
number of cases = single digits + double digits + .. + 7digit numbers
= 7 + 7^2 + 7^3 + ... + 7^7
a geometric series, with a = 7, r = 7 , n = 7
sum = 7(7^7 - 1)/(7-1) = 960799
b) repeats are not allowed
single digits --- 7
double digits, choices = C(7,2) = 21 , arranged in 2! ways = 42
triple digits --- C(7,3) x 3! = 210
4 digits ------- C(7,4) x 4! = 840
5 digits ------- C(7,5) x 5! = 2520
6 digits -------- C(7,6) x 6! = 5040
7 digits --------- 7! = 5040
add them up , notice the above number are just like using P(n,r)
c) can only apply to 3 digit or more numbers.
Let me know your thinking on this.
sorry its supposed to be how many different five digit numbers! can be formed from the seven. sorry!!!!
That makes the whole question much easier!
So what do you think?
for a i had
7 x 7 x 7 x 7 x 7
= 16, 807
for b i had
7 x 6 x 5 x 4 x 3
= 2520
for c i had
7 x 7 x 6 x 7 x 7
c i am very unsure of though.
To find the number of different numbers that can be formed from a given set, we can use the concept of permutations.
a) When repetition is allowed:
In this case, we can choose any of the seven digits (1-7) for each position independently. Since repetition is allowed, we have seven choices for each of the three positions. Therefore, the total number of different numbers that can be formed is 7 x 7 x 7 = 343.
b) When repetition is not allowed:
In this case, once we choose a digit for a particular position, we cannot choose the same digit for any other position. For the first position, we have seven choices (1-7). For the second position, we have six choices remaining, and for the third position, we have five choices remaining. Therefore, the total number of different numbers that can be formed is 7 x 6 x 5 = 210.
c) When repetition is allowed and the third digit must be different from the second digit:
In this case, we need to consider the restrictions for the second and third digits. For the first position, we have seven choices (1-7). For the second position, we still have seven choices, including the possibility of choosing the same digit as the first position. However, for the third position, we need to exclude the possibility of choosing the same digit as the second position. So, we have six choices remaining. Therefore, the total number of different numbers that can be formed is 7 x 7 x 6 = 294.
By following these explanations, you can easily determine the number of different numbers that can be formed in each scenario.