1:There are 7 seats in the family new van. if 4 of the 7 family members can drive,how many possible seating arrengements are there for a family trip in the van?
2:How many even five digit numbers can be formed using only the numbers through 1 and 7?
first the driver in the driver seat
4 possible
that leaves three with the other three to choose the other six seats. How many arrangements of those six people?
the first one has 6 choices, the second 5 choices .....
6*5*4 etc or 6!
so 4 * 6!
even ends in 2 , 4 , 6
so the fifth digit must be 2, 4, or 6
the fourth must be 1, 2 , 3 , 4 , 5 ,6 , 7 seven choices
the third - 7 choices
the second -- 7 choices
he first - 7 choices
so
2 * 7^4
3*7^4
1: To find the number of possible seating arrangements, we can use the concept of permutations. In this case, since there are 7 seats in the van and 4 family members can drive, we need to select 4 people from a group of 7.
The number of ways to select 4 people from a group of 7 is given by the combination formula, which is calculated as:
C(n, r) = n! / (r!(n-r)!)
where C(n, r) represents the combination of selecting r items from a set of n items.
In this case, n = 7 (total number of family members) and r = 4 (number of driving family members). Plugging these values into the combination formula, we get:
C(7, 4) = 7! / (4!(7-4)! )
= 7! / (4!3!)
= (7 * 6 * 5 * 4 * 3 * 2 * 1) / ((4 * 3 * 2 * 1) * (3 * 2 * 1))
= 35
Therefore, there are 35 possible seating arrangements for the family trip in the van.
2: To find the number of even five-digit numbers that can be formed using only the numbers 1 to 7, we need to consider some constraints.
Since the number needs to be even, the units digit must be either 2, 4, 6. The remaining four digits can be filled with any number from 1 to 7.
We have three options for the units digit, and for each of these options, we have 7 choices for each of the remaining four digits.
So, the total number of even five-digit numbers that can be formed is calculated as:
Number of options for units digit * Number of options for each remaining digit
= 3 * 7 * 7 * 7 * 7
= 3 * 7^4
= 3 * 2401
= 7203
Therefore, there are 7203 even five-digit numbers that can be formed using only the numbers from 1 to 7.