A lecture hall contains $20$ chairs, all lined in a row. What is the number of ways that five chairs can be chosen, so that no two chairs are adjacent?

Step-by-step explanation:

If we choose chairs having odd number in the row

no of chairs from which selection is made = 10

no of chairs to be selected = 5

no of ways = 10C₅

similarly if we choose hairs having even numbers only ,

similar to above , no of ways

= 10C₅

Total no of ways

= 2 x 10C₅

= 2 x 10 x 9 x 8 x 7 x 6 / 5 x 4x3 x 2 x 1

= 504 .

4368

(The other answer was wrong)

Call the 5 chairs a, b, c, d, e. Then, since they can't be adjacent, we need at least one chair between them. So it looks something like this

a [] b [] c [] d [] e

We have 15 chairs left (20 - 15 = 5), and we must insert 4, so now there are 15 - 4 = 11 chairs to distribute across six slots:

(here) a (here) b (here) c (here) d (here) e (here)

This is just solving a + b + c + d + e + f = 11 where a, b, c, d, e, f are non-neg ints. This comes out to 16 choose 5, which is 4368.

Also if you need help on AoPS HW use the Message Board first before , Web2.0calc, MSE, etc plz!

Hope that helped,

-sirswagger21

x ≥ 18 is your answer

step-by-step explanation:

let "a number" = x

(2/3)x + 17 ≥ 29

isolate the variable (x). do the opposite of pemdas (parenthesis, exponent (& root), multiplication, division, addition, subtraction).

first, subtract 17 from both sides.

(2/3)x + 17 (-17) ≥ 29 (-17)

(2/3)x ≥ 29 - 17

(2/3)x ≥ 12

next, multiply 3/2 to both sides to isolate the x

(3/2)(2/3)x ≥ 12(3/2)

x ≥ 12(3/2)

simplify. multiply, then divide

x ≥ (12 * 3)/2

x ≥ (36)/2

x ≥ 18

x ≥ 18 is your answer

~

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