Answer:
m = 18
Step-by-step explanation:
To solve this problem, we need to find the value of m such that the number of ways to choose 4 marbles is equal to the number of ways to choose m marbles from a set of 21 marbles.
The number of ways to choose k items from a set of n items is given by the binomial coefficient, also known as "n choose k," which is denoted as C(n, k).
In this case, the number of ways to choose 4 marbles from 21 marbles is C(21, 4), and the number of ways to choose m marbles from the same 21 marbles is C(21, m).
We are given that C(21, 4) = C(21, m).
Using the formula for binomial coefficients, we have:
C(21, 4) = C(21, m)
21! / (4! * (21-4)!) = 21! / (m! * (21-m)!)
Simplifying further:
(21! * m! * (21-m)!) / (4! * (21-4)!) = 1
Cancelling out the common terms:
(m! * (21-m)!) / (4! * (21-4)!) = 1
Simplifying the factorials:
(m! * (21-m)!) / (4! * 17!) = 1
(m! * (21-m)!) = (4! * 17!)
Since factorials are always positive, we can remove the factorials from both sides:
(m * (m-1) * ... * 1) * ((21-m) * (21-m-1) * ... * 1) = (4 * 3 * 2 * 1) * (17 * 16 * ... * 1)
Cancelling out the common terms:
(m * (m-1) * ... * 1) * ((21-m) * (21-m-1) * ... * 1) = (4 * 3 * 2 * 1) * (17 * 16 * ... * 1)
Expanding the products:
m! * (21-m)! = 24 * 17!
We know that 24 = 4 * 6, so we can rewrite the equation as:
m! * (21-m)! = (4 * 6) * 17!
We see that 6 is a factor in both m! and (21-m)!, so we can simplify further:
(6 * (m! / 6) * ((21-m)! / 6)) = 4 * 17!
Simplifying:
(m-1)! * ((21-m)! / 6) = 4 * 17!
Since 17! does not have a factor of 6, we know that (21-m)! / 6 must equal 1:
(21-m)! / 6 = 1
Solving for (21-m)!, we have:
(21-m)! = 6
The only positive integer value of (21-m)! that equals 6 is (21-m)! = 3.
Therefore, (21-m) = 3, and solving for m:
21 - m = 3
m = 21 - 3
m = 18
Thus, the value of m is 18.