What is the 4th term in the binomial expansion of (a+b)^6)?

Mathematics

1 answer:

Sergeeva-Olga [200]4 years ago

8 0

Answer:

option A) that is 20a^3 b^3

Step-by-step explanation:

Given in the question the equation,

(a+b)^6

Formula for binomial expansion up to 4th term of (a+b)^6)

1st term 2nd term 3rd term 4th term

6C0 (a)^6(b)^0 + 6C1 (a)^5(b)^1 + 6C2 (a)^4(b)^2 + 6C3 (a)^3(b)^3 + ....

Hence, the 4th term in the binomial expansion of (a+b)^6 = 6C3 (a)^3(b)^3

= 20 (a)^3(b)^3

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you have enough tickets to play 6 different games at a amusement park. if there are 14 games, how many ways can you choose six?

Musya8 [376]

Answer:

3003 ways

Step-by-step explanation:

You can basically choose 6 games from 14 games in total. This is essential a combination problem. We want the number of ways to choose 6 things from 14 things. The general formula for combinations is:

$nCr=\frac{n!}{r!(n-r)!}$

Which tells us the number of ways to choose "r" things from a total of "n" things.

The factorial notation is:

n! = n * (n-1) * (n-2) * ....

Example: 3! = 3 * 2 * 1

Now, we know from the problem,

n = 14

r = 6

So, substituting, we get:

$nCr=\frac{n!}{r!(n-r)!}\\14C6=\frac{14!}{6!(14-6)!}\\=\frac{14!}{8!*6!}\\=\frac{14*13*12*11*10*9*8!}{6!*8!}\\=\frac{14*13*12*11*10*9}{6*5*4*3*2*1}\\=3003$