Question

$10(x4 \times 5) + 55$
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Answer 1

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$10(x4 \times 5) + 55$

[tex]40x \times 50 + 55 \\ 2000x + 55

Hope this will be helpful.

Answer 2

Answer:

Step-by-step explanation:

6.5: 10, 11, 16, 18, 20, 24, 26x, 47.

10 (a) Since there are 6 kinds of croissant, this is the same as counting the number of non-negative integer

solutions to x1 + x2 + x3 + x4 + x5 + x6 = 12. That is, there are C(12 + 6 − 1, 6 − 1) = C(17, 5) possbile

choices.

(b) Same as before, but for solutions to x1 + x2 + x3 + x4 + x5 + x6 = 36, of which there are C(41, 5).

(c) Now we count the number of solutions to x1 + x2 + x3 + x4 + x5 + x6 = 24 where xi ≥ 2. Letting

yi = xi −2, this is the same as counting non-negative integer solutions to y1 +y2 +y3 +y4 +y5 +y6 = 12.

So the answer is C(17, 5).

(d) Now we count solutions to x1 + x2 + x3 + x4 + x5 + x6 = 24 where x6 ≤ 2. We do this by counting all

non-negative solutions and then throwing away (subtracting) the solutions for which x6 > 2. In detail:

There are as many solutions to x1+x2+x3+x4+x5+x6 = 24 with x6 > 2 as solutions to x1+x2+x3+

x4 +x5 +x6 = 22 with xi ≥ 0, that is C(27, 5). We subtract this from C(29, 5) to get C(29, 5)−C(27, 5)

ways to choose two dozen croissants with no more than two broccoli.

(Note: It is also possible to count and add the possibilities with no, one and two broccoli croissants.

This would be just slightly more work and should give the same answer.)

(e) Now we count solutions to x1 +x2 +x3 +x4 +x5 +x6 = 24 with x3 ≥ 5 and x4 ≥ 3. Let y3 = x3 −5 and

y4 = x4 − 3. Then this count is the same as counting the solutions to x1 + x2 + y3 + y4 + x5 + x6 = 16

with non-negative numbers. So, C(21, 5).

(f) This is equivalent to counting solutions to

x1 + x2 + x3 + x4 + x5 + x6 = 24 − 1 − 2 − 3 − 1 − 2 = 15

where x6 ≤ 3.

There are C(20, 5) to this equation without the constraint, and there are as many solutions with x6 ≥ 4

as there are unconstrained solutions to

x1 + x2 + x3 + x4 + x5 + x6 = 11,