The number of words that can be formed without restriction are (8!/2!2!). Cases where the two Is are always together are 7!/2! cases. Hence the number of words where they are never together are 8!/2!2! - 7!/2! = 7560.

Let the number of black, blue and burgundy balls be a, b, and c. Total number of ways will be the total number of solutions of the equation a+b+c=25
No. of ways= $$^{25+3-1}C_{3-1}=^{27}C_{2}=351$$ ways.

Let's assume the machine produces 1000 bulbs in a day. Then the number of bulbs produced by P, Q and R is 250, 450 and 300 respectively.
Since the defective percentage of P, Q and R are given as 4%, 6% and 5%, the number of defective bulbs produced is 10, 27 and 15 respectively.
Thus, the total number of defective bulbs in the day = 52 out of which 10 are produced by P.
Thus, the required probability = 10/52 = 5/26

or By Bayes Theorem,

P(P/Defective) = $$\frac{P(D/P)\times P(P)}{P(D/P)\times P(P) + P(D/not P)P(not P)}$$

= $$\frac{10/250 \times 250/1000}{(10/250 \times 250/1000) + (750/1000 \times 42/750)} $$

= 5/26

p+q+r+s+t = 50
q gets a minimum of 4 => after giving 4 to q we get p+q+r+s+t = 46 => $$^{50}C_4$$
Now we have to remove the cases where t has more than 7.
So after giving T 8 perks, we get p+q+r+s+t = 38 => $$^{42}C_4$$
=> $$^{50}C_4$$ - $$^{42}C_4$$ = 230300 - 111930 = 118370.

Total number of rectangles on a square board having distinct length and breadth = 9C2*9C2 - ($$1^2+2^2+....+8^2$$) = (36*36) - (8*9*17/6) = 1296 - 204 = 1092.
Now we want the are 18 and dimensions 6 and 3.
Let the breadth be 3 units. If we consider first 3 rows as a breadth, then the total number of rectangles formed is 3.
The breadth of 3 units can be taken in 6 ways.
Total number of rectangles formed = 6*3 = 18

Similarly if we consider length 3 units, Total number of rectangles formed = 18

The number of rectangles with area 18 sq units with one side as 6 units = 18+18 = 36

Probability = 36/1092 = 3/91.

Every 5 balls can be arranged in the ascending order in only one way. So, in a group of 5 balls, there is only one unique way to arrange them in ascending order.
=> Out of 5! Ways to arrange the selected balls, only one is in ascending order.
=> Probablilty= 1/5!=1/120

Total number of rectangles on a square board having distinct length and breadth = 9C2*9C2 - ($$1^2+2^2+....+8^2$$) = (36*36) - (8*9*17/6) = 1296 - 204 = 1092.
Now we want the are 18 and dimensions 6 and 3.
Let the breadth be 3 units. If we consider first 3 rows as a breadth, then the total number of rectangles formed is 3.
The breadth of 3 units can be taken in 6 ways.
Total number of rectangles formed = 6*3 = 18

Similarly if we consider length 3 units, Total number of rectangles formed = 18

The number of rectangles with area 18 sq units with one side as 6 units = 18+18 = 36

Probability = 36/1092 = 3/91.

The area of the total possible solutions is the square with corners at (0,0); (0,4); (4,0) and (4,4). The required area is given by the area of the triangle X+Y > 6. So, probability is (1/2)* 2* 2/(4*4) = 1/8.

We can graphically represent the region from the following figure.

The area of the square (0,0),(4,0),(4,4),(0,4) is the all probable outcomes.  The area of the triangle (2,4)(4,4)(4,2) is the expected probable region.

0.5x2x2/4x4= 1/8

Arranging EXEMPLARY in alphabetical order we get AEELMPRXY. The first words of the 7 letter list would start from A.

For A, removing the extra E, $$^7 P _6$$ words can be formed from the distinct letters. Now let us take the case when both E are present in the mix. So, out of the 7 letters, 3 letters are A, and 2 Es. So, the other 4 letters can be selected as $$^6C_4$$. So, the words can be arranged in $$^6 C _4 \times (6!/2!)$$ ways. Hence there are $$^7 P _6 + ^6 C _4*(6!/2!)$$ 7-letter words starting with A.
Next in line would be words starting with E.
Let us first consider words of form, EA _ _ _ _ _.
The 5 digits can be arranged in $$^7P_5 ways$$.

Same number of words will be there of the form, EE _ _ _ _ _, EL _ _ _ _ _, EM _ _ _ _ _, EP _ _ _ _ _ , and ER _ _ _ _ _.

Each of these letters produces $$^7 P _5$$ words.
By this way we get that rank= $$(^7 P _6 + ^6 C _4 * (6!/2!))+(6* ^7 P _5) +(2 * ^5 P _3)+(2* ^4 P _2) + (1* ^3 P _1) +1 = 25708$$.

Two black squares can be chosen in $$ ^{32} C _2$$ ways.
The number of ways we can choose two black squares that lie on the same row is $$(^{32} C _1 * ^3 C _1)/2$$.
Similarly, for column.
Hence, ways to select black squares that don't lie on the same row or column is $$ ^{32} C _2 - ^{32} C _1 * ^3 C _1 = 400$$.

The numbers that give the same units digit on being squared are 0, 1, 5, 6. Hence the chance of getting such a number in units digit is 2/5.

After the first ball is chosen, there would be 9-1=8 balls with that unit digit remaining out of 89. Hence, the probability that the second number has the same units digit = 8/89

Hence p = 2/5 * 8/89 = 16/445

The letters can be picked from a set of 26. Therefore, the number of ways of selecting them = 26*26*26. A non-ascending or non-increasing set of numbers means that the numbers to the right are less than or equal to the numbers on the right.

For example, 987 is a decreasing sequence of numbers while 997 is a non-increasing (non-ascending) set of numbers. Please note the difference between non-ascending and not ascending.

In order to select three numbers in a non-ascending sequence, if all three of them are different, the number of cases is 10C3. Note that in this case, the number is non-ascending as well as descending.

If exactly 2 of the three numbers are same, the number of ways selecting the two numbers is 10C2. Once the two numbers are selected, the number of ways of arranging them is 2. So, total is 10C2*2 = 90. Note, that numbers like 221,211 fall in this category. These are not descending numbers.

The last case if when all the three numbers are the same. Number of ways is 10.

So, total is 120+90+10 = 220.

Hence, number of registration numbers possible = $$26^3*220.$$

If the sum is less than 7, it can either be 2, 3, 4, 5, or 6. 

For a sum of 2, the possible outcomes are (1,1).

For a sum of 3, the possible outcomes are (1,2) and (2,1).

For a sum of 4. the possible outcomes are (1,3), (3,1), (2,2).

For a sum of 5. the possible outcomes are (1,4), (4,1), (2,3) and (3,2).

For a sum of 6, the possible outcomes are (1,5), (5,1), (2,4), (4,2), (3,3).

The total number of favorable outcomes = 15

The total number of outcomes = 36

Hence, the probability = 15/36 = 5/12.

Consider 1,2,3,4,5 as one unit. Now 1 and 5 have to be at extreme ends.

Number of ways of arranging this group = 2*3!

Number of ways of ways of arranging this group and beads 6, 7, 8, 9, 10 = 5!*2*3! = 12*5!

But we have to build a necklace from these beads.

As there is no difference between clockwise and anticlockwise arrangements, we have to divide the number of arrangements by 2.

Number of ways of building a necklace = 6*5! = 6!

The total number of solutions without the restriction of a>b is $$^{13}C_3 = 286$$

Let us now calculate the number of solutions where a=b.

When a=b=0, the equation becomes c+d = 10 -> No. of solutions = 11

When a=b=1, the equation becomes c+d = 8 -> No. of solutions = 9

And so on.

So the number of solutions where a=b and a+b+c+d=10 is 11+9+7+5+3+1 = 36

So, by symmetry, the number of solutions where a>b and a+b+c+d=10 is (286-36)/2 = 125

Triangles can be formed by picking two points on either of the line and a third point on the other line. This can be done in $$^6 C _2 * ^{11} C _1 + ^{11} C _2 * ^6 C _1 = 495$$ ways.

Case 1. Two persons  go to 1 movie and the rest must go to one movie each.

Number of ways 2 people can go to 1 of the 4 movies = Choice of 2 people * Choice of movie = $$^5 C _2$$ * $$^4 C _1$$

The remaining 3 people can watch the remaining 3 movies in 3! ways.

Number of ways = $$^5C_2*^4C_1*3!$$ = 240.

Case 2. One person does not watch the movie.

 4 people can watch the 4 movies in 4! ways.

Number of ways 4 people can go to the 4 movies = Choice of 4 people * Arrangement of movie among them= $$^5 C _4$$ * $$4!$$

=120 ways.

Total number of ways= 240+120= 360 ways.

Let “T” be the event of speaking the truth and “F” be the event of lying.

P(T) = 1/4, P(F) = 3/4

Let “E” be the event of reporting a six and B be the event of actually getting a six. B' is the event of not getting a six

P(E|T) indicates that a man reports a six after he speaks the truth. This means that dice shows a six

P(E|T) = P(E|T,B)*P(B) + P(E|T,B')*P(B') = 1*1/6 + 0*5/6 = 1/6

P(E|T,B) = Probability of reporting a six if a man wants to tell the truth and six has actually occured = 1

P(E|T,B') = Probability of reporting a six if a man wants to tell the truth and six has actually not occured = 0

Similarly P(E|F) indicates that a man reports a six after he lies. This means that dice does not show six

P(E|F) = P(E|F,B)*P(B) + P(E|F,B')*P(B') = 0*1/6 + 1/5*5/6 = 1/6

P(E|F,B) = Probability of reporting a six if a man wants to tell a lie and six has actually occured = 0

P(E|F,B') = Probability of reporting a six if a man wants to tell a lie and six has actually not occured = 1/5*5/6

We multiplied 1/5 as there are 5 possibilities of speaking a lie if 6 does not come and out of 5 possibilities, 6 is chosen

We have to find P(T|E)

Applying Bayes Theorem

P(T|E) =$$\frac{P(T)P(E|T)}{P(T)P(E|T)+P(F)P(E|F)}$$

P(T|E) = $$\frac{1/4 * 1/6}{1/4*1/6+3/4*5/6*1/5} = 1/4$$

Let us consider the number of ways in which we can reach the nth step.
We can reach the nth step by

• Reaching n-1th step and then take 1 step to reach n
• Reach n-2th step and take 2 steps to reach n
• Reach n-3th step and take 3 steps to reach n

Hence the number of ways of reaching the nth step will be the sum total of the number of ways of reaching the (n-1)th, (n-2)th and (n-3)th steps. Note that there are no overlaps in the above three methods, as the number of steps take in the last differ.
Hence we have $$T_{n} = T_{n-1}+T_{n-2}+T_{n-3}$$

1st step can be climbed in 1 way
2nd step can be climbed in 2 ways (11, 2)
3rd step can be climbed in the form of 111,12, 21, 3 => 4 ways
We can apply the above formula from here.
4th step can be climbed in 1 + 2 + 4 = 7 ways
5th step can be cimbed in 2 + 4 + 7 = 13 ways
6th step can be climbed in 13+7+4 = 24 ways
7th step can be climbed in 24+13+7 = 44 ways
8th step can be climbed in 44+24+13 = 81 ways
9th step can be climbed in 81+44+24 = 149 ways
10th step can be climbed in 149+81+44 = 274 ways
11th step can be climbed in 274+149+81 = 504 ways
12th step can be climbed in 504+274+149 = 927 ways

There are 6 steps in total out of whih 3 are horizontal and 3 are vertical:HHHVVV.
Number of ways of reaching origin is = 6!/(3!*3!) = 20

There are 6 steps in total out of whih 3 are horizontal and 3 are vertical:HHHVVV.
Number of ways of reaching origin is = 6!/(3!*3!) = 20

The number of notes taken out = 9x
The maximum number of notes withdrawn such that no denomination has 8 notes will be possible when we withdraw 7 notes of all the 5 denomination. Now if we draw even a single note after this, at least one denomination will have 8 notes. 

Hence the required number of notes  = 7*5+1=36
Hence the minimum number of boxes = 36/9 = 4.

A number is an even steven number if all its digits are even.
The number can have a maximum of six digits and each of these digits can be selected in 5 ways.
Note that 0 can be any of the digits.
Hence the number of numbers is $$5^6 = 15625$$.