practice problems pending

1. Find the total number of words formed by the letters a,a,b,b,c.
S1. 5!/2!.2!

2. In how many ways can you arrange 2 beads of red colour, 2 beads of blue colour and 3 beads of black colour in a ring? Note that beads with the same colour are identical and arrangements that can be obtained by rotating (or) flipping the ring are identical.
S2.
First attempt(wrong):
(7-1)! circular permutations
Divide by 2!2!3! to account for repeated elements.
Divide by 2 to account for flipping.
Answer = 15

This is wrong, correct answer is 18.
Why?

Let's see using a smaller example.
2 blue, 3 black.
(5-1)! circular permutations
Divide by 2!3!.
So far so good.
Mistake is when we divide by 2 for flipping.

Why do we even divide by 2 for flipping.

Let's say we have 4 distinct beads.

Here is one necklace:

1 --- 2
|       |
4 -- 3

and flipping it we get:
2 --- 1
|       |
3 -- 4

So 1-2-3-4 was clockwise and it became counter clockwise.
Flipping gives a distinct permutation.
Hence we divide by 2.

But what if the flipping gave the same permutation back?

Let's consider the case with 2 blue and 3 black beads.
Blue = B, Black = K.

Our approach gives.
1. (5-1)!
2. Divide by 2!3!
3. Divide by 2

Answer = 1 which is wrong correct answer = 2.
Let's see why.

Our solution till step 2 is correct.
We have 2 till step 2.

Let's see which 2 permutations they are:
BBKKK -ie blue together black together
BKBKK - ie 2 blue separated by 1 black.

         |

        (K)

      /  |  \

   K   |   K

    \   |   /

   B--|--(B)

        |

         |

        (K)

      /  |  \

   B   |   B

    \   |   /

   K--|--K

        |

If we flip them around the vertical axis as shown, we are creating the same necklace back.
So we don't have to divide by 2, since we didn't create a new pattern.

Similar logic goes for the original problem.
Until the second step we had 30 permutations.
Now we have to figure out how many of those will give the same necklace back upon flipping.
And those permutations we won't divide by 2.
Since there are 3 black beads and both others are 2 each, we have to flip it across an axis passing through the black one.

      K
B           B
R           R
K           K

For e.g. the above permutation upon flipping will remain the same.
How many such are there?
Just look at LHS.
There are 3! ways to arrange LHS = 6.
So 30 - 6 = 24 are the ones giving distinct permutations upon flipping.
So answer = 24/2 + 6 = 18.

3. Let S be the set of permutations of the sequence 1, 2, 3, 4, 5 for which the first term is not 3. A permutation is chosen randomly from S. The probability that the second term is 1, in lowest terms, is a/b. What is a + b?
S3.
|S| = 4.4.3.2 = 96
Now second term is 1 and first term can't be 3: 3.1.3.2.1 = 18
a/b = 18/96 = 3/16
a+b = 19

4. A positive integer divisor of 12! is chosen at random. The probability that the divisor chosen is a perfect square can be expressed as m/n where m and n are relatively prime positive integers. What is m+n?

S4.

12! = 2^10.3^5.5^2.7.11

Total divisors = 11.6.3.2.2

For squares, look at it as 25^1 * 9^2 * 4^5

Square divisors = 2 * 3 * 6 = 36

m/n = 36/26.22 = 1/22

m +n = 23

5. How many four-digit positive integers have at least one digit that is a 5 (or) a 8?

S5.
Total: 9*10*10*10
No 5 or 8 => Remaining 8 digits only => 7 * 8 * 8 * 8
Answer = 9000 - 3584 = 5416