Question

in how many ways 4 people be selected at random from 6 boys and 4 girls if there are exactly 2 girls​

Answer 1

Step-by-step explanation:

NO. OF BOYS = 6

NO. OF GIRLS = 4

Exactly 2 girls are there

$seection \: \: of \: \: girls = 4c_{2} \\ = \frac{4!}{2! \times (4 - 2)!} \\ \\ = \frac{4 \times 3 \times 2 \times 1}{2 \times 1 \times 2 \times 1} = 6$

Now 2 girls are fixed then only two members are left

$selection \: \: of \: \: boy \: \: = 6 c_{2} \\ = \frac{6!}{2! \times (6 - 2)!} \\ \\ = \frac{6 \times 5 \times 4 \times 3 \times 2 \times 1}{2 \times 1 \times 4 \times 2 \times 3 \times 1} \\ \\ = 15$

Now

$no \: \: of \: \: selection \: \: = \: 6 \times 15 \\ = 90$

Answer 2

There are 90 ways.

• Combinations are mathematical operations that count the number of potential configurations for a set of elements when the order of the selection is irrelevant. You can choose the components of combos in any order. Permutations and combinations can be mixed up. The sequence in which the chosen components are chosen is crucial in permutations, too. For instance, the arrangements ab and ba are distinct in permutations but equivalent in combinations (regarded as one arrangement).
• Combinatorics is the study of combinations, but mathematics and finance are two other fields that use combinations.

Here, according to the given information,

There are 6 boys and 4 girls.

Now to select 2 girls from 4 girls, we get,

$4C_{2} = 6$.

Again, since 2 girls have been taken, selecting 2 boys out of 6 boys, we get,

$6C_{2} = 15.$

Hence, Number of ways 4 people be selected at random from 6 boys and 4 girls = 6 × 15 = 90.

Hence, there are 90 ways.

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