Question

For each of the following scenarios, match to the applicable formula and explain why it fits with that formula.

James wants to see how many ways he can select some poker chips if he has 4 red, 2 blue and a black. How many ways can he place a bet?

James decides to put his 4 red, 2 blue and 1 black poker chip in a line. How many orders can he do this in?

James decides he will trade his poker chips so he only has one of each colour (blue, red and black) but he gets a green one as well. How many ways can he place a bet now?

Out of his four poker chip, James decides to select two to bet. How many different ways can he do this?

James wants to place 3 bets of one chip each in a game. In how many orders can he accomplish this with his four chips?

a. n!/a!b!c!...

b. n P r

c. n C r

d. (p+1)(q+1)(r+1)... - 1

e. 2^n - 1

Answer 1

James has 4 red(R), 2 blue(B) and 1 black(b) chip.

How many ways can he place a bet?

James has 4R, 2B, 1b

Now he can select either 0R, 1R, 2R, 3R or 4R => Total ways = 5 = (4+1)

Similarly, he can select 0B, 1B or 2B = > 3 ways = (2+1)

And he can select 0b or 1b => 2 ways = (1+1)

Hence total ways of selections = (4+1)*(2+1)*(1+1)

But to place a bet, there should be at least one chip. So subtract the case when he selected 0R, 0B, 0b

Hence Total ways of making a bet = (4+1)*(2+1)*(1+1) - 1

d. (p+1)(q+1)(r+1)... - 1 is the correct match for this.

James decides to put his 4 red, 2 blue, and 1 black poker chip in a line. How many orders can he do this in?

Note that chips of the same color are identical. If chips were different then the answer would have been n! directly.

But due to identical objects, it would be different. Suppose we need to arrange A1, A2, B where A1 and A2 are identical so the arrangement A1A2B and A2A1B are the same. Both arrangements would be counted as one only due to indistinguishable objects. So counting of such things is given by-  (total)! / (n1)! (n2)1 .....

Hence the number of orders he can do is 7! / 4!2!1!

a. n!/a!b!c!... is the correct match.

James decides he will trade his poker chips so he only has one of each colour (blue, red and black) but he gets a green one as well.

Now James has this situation- 1R, 1B, 1b, 1G and the previous situation is no more.

How many ways can he place a bet now?

Now for each chip, he has two options, either take the chip or reject. So total ways = 2*2*2*2.

But to place a bet, there should be atleast one chip. So subtracting the case when he rejected all the chips

Total ways to place a bet = 2*2*2*2 - 1

e. 2^n - 1 is the correct match

Out of his four poker chip, James decides to select two to bet. How many different ways can he do this?

He has 1R, 1B, 1b, 1G, all different.

So the number of ways of selecting r objects from n different objects is nCr

4C2 is the answer.

c. nCr is the correct choice

James wants to place 3 bets of one chip each in a game. In how many orders can he accomplish this with his four chips?

Let's suppose James choose 3 out of 4 chips to bet, no. of ways of selecting = 4C3

Also, it is uncertain which chip he will choose first in the game.

Total arrangements of chips = 3! because 3 places 3 bets of one chip each.

Hence orders he can accomplish = 4C3*3! = 4P3

b. nPr is the correct choice.