The symbol nCr, rea "n choose r" or "the number of combinations of n things taken r at a time" is the number:
n!
-----
r!(n-r)!
a:find 4C1
b:Find 5C0 and 5C5
c: Find 7C1, 7C2, 7C3
What is your question on this? The formula gives the algorithm.
a)4!/1!(4-1)!=4 x 3 x 2/3 x 2, cancel out the 3's and 2's=4
You do b and c. Post your answers if you'd like them checked.
To find the value of nCr, where n is the total number of items and r is the number of items chosen at a time, we can use the formula:
nCr = n! / (r!(n-r)!)
Now let's solve the given problems:
a) 4C1 = ?
Plugging in the values into the formula, we have:
4! / (1!(4-1)!) = 4! / (1! * 3!) = (4 * 3 * 2 * 1) / (1 * 3 * 2 * 1) = 4
So, 4C1 is equal to 4.
b) 5C0 = ? and 5C5 = ?
For 5C0, we have:
5! / (0!(5-0)!) = 5! / (1 * 5!) = 1
Hence, 5C0 is equal to 1.
For 5C5, we have:
5! / (5!(5-5)!) = 5! / (5! * 0!) = 1
Therefore, 5C5 is also equal to 1.
c) 7C1, 7C2, and 7C3 = ?
For 7C1:
7! / (1!(7-1)!) = 7! / (1! * 6!) = 7
So, 7C1 is equal to 7.
For 7C2:
7! / (2!(7-2)!) = 7! / (2! * 5!) = 7 * 6 / 2 = 21
Hence, 7C2 is equal to 21.
For 7C3:
7! / (3!(7-3)!) = 7! / (3! * 4!) = (7 * 6 * 5) / (3 * 2 * 1) = 35
Therefore, 7C3 is equal to 35.
In summary:
a) 4C1 = 4
b) 5C0 = 1 and 5C5 = 1
c) 7C1 = 7, 7C2 = 21, and 7C3 = 35