The probability of Jack picking a yellow or white sock is 0.6 (60%).

The probability of Jack picking a blue sock is 4/14 (28.6%)

The probability of Jack picking a red sock is 3/14 (21.4%)

The probability of Jack picking a white sock is 2/14 (14.3%)

The probability of Jack picking a yellow sock is 1/14 (7.1%)

Therefore, the probability of Jack picking a yellow or white sock is 0.6 (60%)

Yes very fun and totally not pain and suffering. 🙂

@smilingtadpole-1 In this problem, we do not care whether 11 has one or two 1s. What matters is the fact that 11 contains a 1. When you solved this problem, you assumed that because 11 had two ones, it should be counted twice, which is incorrect. I hope this helps!(also charge your device lol)

Thanks for the explanation. I guess I didn’t think that the card had numbers on both sides. But it makes sense now.

-1x-2+3-4+(5x6-7^2)- -18=?

@versatilemole i could give as many solutions as i can to this problem, and then people can add on solutions and count how many ways we can make 100.

Not really 🙂 Notice how Elmo has three blocks. It wouldn't be $$\binom{9}{3}$$ because you use a separate block every time so you would have to choose the first block, then the second, and so on. Try to see where you can go from there 😉

Note: Sorry this is so late I doubt you're active on this forum anymore.

@knowledgeabledog Thanks!

@rationaltiger Correct!

=1+11+111 times 68
=12+548
=560

24+3/4(24)(2)+(3/4^2)(24)(2)+(3/4^3)(24)(2)+.....
(The x2 is because the ball drops the same distance up and down)

This is an infinite arithmetic sequence.

Let S = 3/4(24)(2)+(3/4^2)(24)(2)+... = 36+(3/4)(36)+(3/4^2)(36)+....
So 4/3S=48+36+(3/4)(36)+....
So 4/3S-S=1/3S=48
So S=144
So S+24=168?

MISSPELL has 8 letters with 2 S’s and two L’s. 8!/(4!2!) = 10,080 arrangements, but one of these is the correct spelling, so there are 10,079 misspellings in total.