# Combinations MCQ 1( bag and balls )

Combination MCQ 1: A bag contains nine yellow balls, three white balls and four red balls. In how many ways can two balls be drawn from the bag?

Solution:

Formula for combinations of n things with chosen  r at a time is

$_n\rm C_2= \frac {n!}{r!(n-r)!}$

Here we have;

n = 9+3+4=16 (total balls)

$=\frac{16\times 15\times 14\times 13\times 12\times 11\times 10\times 9\times 8\times 7\times 6\times 5\times 4\times 3\times 1}{2\times 1(14\times 13\times 12\times 11\times 10\times 9\times 8\times 7\times 6\times 5\times 4\times 3\times 2\times 1)}$

r=2 ( two ball from among 16 balls to be chosen)

So, by above formula, the number of combinations in which we can take  two balls are;

$_16\rm C_2= \frac{16!}{2!(16-2)!}$

We know;

16! = 16x15x14x13x12x11x10x9x8x7x6x5x4x3x1

2!=2×1

(16-2)!=14!=14x13x12x11x10x9x8x7x6x5x4x3x2x1

putting these values in above formula;

$= \frac{16 \times 15\times 14\times 13\times 12\times 11\times 10\times 9\times 8\times 7\times 6\times 5\times 4\times 3\times 1}{2\times 1 (14\times 13\times 12\times 11\times 10\times 9\times 8\times 7\times 6\times 5\times 4\times 3\times 1)}$

common terms (as red color) in this fraction will cancel out — we then get;

$=\frac{16 \times 15}{2 \times 1}=120$

So, we can choose two balls from given in 120 different ways or schemes !

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