If we will chose two kings from the deck of 52 cards, in how many ways we can select three more cards from the same deck to have three of the same kind and two of another.

192 total ways

Step-by-step explanation:

This time, we have already 2 selected kings, therefore we must find the remaining 3 cards.

There are two general ways to do this:

Case 1: you draw two of some other matching cards and another king

Case 1 - K + X + X where K is a King and X is some other card than a King. So there are 2 Ks to draw. Then there are 12 other cards that are not from the king, and since you get two of them it would be 2 taken from 4:

4C2 = 4! / (2! * (4-2)!) = 6

Therefore there are 6 different arrangements of the two.

So 2 * 12 * 6 = 144 total ways to do this.

Case 2: you draw three of some other matching cards

Case 2 - X + X + X where the Xs are all non-king cards. So 12 cards to choose from, and then calculate 3 taken from 4:

4C3 = 4! / (3! * (4-3)!) = 4

Therefore there are 4 different arrangements of the three. So 12 * 4 = 48 total ways to do this.

Which means that:

144 + 48 = 192

There are 192 total ways to get a full house when you start with two kings.

add all the numbers and check the ones place.

the answer is 486. as the ones place is more than 4, the estimation will be the next 10 ie, 490

given hc ∥ gd.¯¯¯¯¯¯which statement is true

step-by-step explanation:

∠cbe and ∠deb are same-side interior angles