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Productfor(n, 10, 29, (n+1) / n)==11/10 * 12/11 * 13/12................30/29==3

The first fraction==11 / 10

I am having same problem. Any proper solution ?

sorry you are wrong

y = 4x - 13/5

What is the "common difference" of the sequence you have written down: 94, 87, 73, 66.....etc. Can you tell me?

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What is the definition, or the formula for "expected value" in your Math Textbook?

YES! There is a counting method oulined here in Wikipedia:  https://en.wikipedia.org/wiki/Twelvefold_way

Look at the table at the very bottom of the page. EVERY case of distributions of balls into boxes (or whatever the case may be) is there with the CORRECT formula to be used. Good luck.

Where are the NUMBERS in your question?  " He has ? fractions, and their product equals ? "  ??!

10 - 99 : 11
100 - 999 : 112
1000 - 9999 : 1124
10000 - 99999 : 11248
100000 - 999999 : 112496
1000000 - 9999999 : 1124992
10000000 - 99999999 : 11249985
100000000 - 999999999 : 112499976
1000000000 - 9999999999 : 1124999972

Why are you stuck? Can you tell me what are the squares and cubes between 1 and 10?

(a)

20 IDENTICAL stickers to 5 DIFFERENT friends:

[26 + 5 - 1] nCr 5 ==142,506 ways - with no restrictions.

(b)  Same as (a) above except everybody get " at least one sticker"

(26 - 1) nCr (5 - 1) ==25 nCr 4 ==12,650 ways

SOURCE:  https://en.wikipedia.org/wiki/Twelvefold_way

{See the table at the very bottom of the page. Item #7 for (a), item #9 for (b)}

Where are the NUMBERS in your question? " The resulting product is less than ......?"

Where are the NUMBERS in your question?  "  If there are skaters, including Americans" ?? 

Balls are DISTINCT and boxes are IDENTICAL:

sumfor(k, 0, 3, (5 nCr k)) ==26 - with no restrictions.

Source:  https://en.wikipedia.org/wiki/Twelvefold_way

{See the table at the very bottom of the page - item # 1]

What's the f ?

1.4268292682926829^(1/20) or that^0.05 = 1.0179316092382140458823473304434

Scientific calculators go down to the last decimal... No need to post this question because this website has a literal calculator :O

Yes there definitely is a good strategy. We have 10 different variables.

Case 1, the sum is = 0:

There is 1 solution, where all variables are equal to 0.

Case 2, the sum is = 1:

There are 10 choose 1 ways to arrange the value of 1 among the 10 variables, so there are 10 solutions.

Case 3, the sum is = 2.

We can have 10 ways to arrange the 2. But if we have two 1s, we have (10 choose 2) ways to do it = 45.

Total sum is 10 + 45 + 10 + 1 = 66 total cases.