11c2
11c2 can be expanded using the binomial theorem.
The binomial theorem states that for any positive integer n and any real numbers a and b, the expansion of (a + b)^n is given by:
(a + b)^n = C(n, 0) * a^n * b^0 + C(n, 1) * a^(n-1) * b^1 + C(n, 2) * a^(n-2) * b^2 + ... + C(n, n-1) * a^1 * b^(n-1) + C(n, n) * a^0 * b^n
Where C(n, k) denotes the binomial coefficient, which is given by:
C(n, k) = n! / (k! * (n - k)!)
In this case, n = 11 and k = 2. Plugging these values into the binomial theorem, we get:
(11c2) = C(11, 0) * 11^11 * 2^0 + C(11, 1) * 11^10 * 2^1 + C(11, 2) * 11^9 * 2^2 + ... + C(11, 9) * 11^2 * 2^9 + C(11, 10) * 11^1 * 2^10 + C(11, 11) * 11^0 * 2^11
Evaluating this expression will give you the value of 11c2.
To calculate 11c2, we need to use the combination formula:
nCr = n! / r!(n-r)!
In this case, n = 11 and r = 2. Substituting these values into the formula:
11c2 = 11! / 2!(11-2)!
Simplifying the expression:
11c2 = 11! / 2!9!
Now, calculating the factorials:
11! = 11 × 10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1
2! = 2 × 1
9! = 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1
Plugging in the values:
11c2 = (11 × 10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1) / (2 × 1)(9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1)
Simplifying further:
11c2 = (11 × 10) / (2 × 1)
11c2 = 55 / 2
Therefore, 11c2 equals 55 divided by 2, which is equal to 27.5.
To calculate 11 choose 2 (11C2), you need to use the combination formula or the binomial coefficient formula, which is:
nCr = (n!)/(r!(n-r)!)
In this case, n = 11 and r = 2.
First, let's find the value of n! (11 factorial):
n! = 11 × 10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1 = 39,916,800
Next, let's find the value of r! (2 factorial):
r! = 2 × 1 = 2
Lastly, let's find the value of (n-r)! ((11-2) factorial):
(n-r)! = (11-2)! = 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1 = 362,880
Now, substitute these values into the combination formula:
11C2 = (11!)/(2!(11-2)!)
= 39,916,800/(2 × 362,880)
= 39,916,800/725,760
= 55
Therefore, 11 choose 2 (11C2) equals 55.