Test: Problem Solving- 3 - UPSC MCQ

• There are 6 students standing in a circle.

• The ball can be passed from one student to any other student except themselves.

• So, from each student, the ball can be passed to 5 other students.

To find the total number of possible passes in one move:

• Number of students = 6

• Number of choices for passing the ball from one student = 5

Total possible passes = 6 x 5 = 30

So, the answer is option B: 30.

So that gives 5x4x3x2 = 120 possible ways to pair up 5 boys and 5 girls into boy-girl couples. Putting the boys in a fixed row, any permutation of the girls will make a valid pairing. So in total there will be 5!= 120 different ways to arrange boy-girl pairs.

Step 1: Arrangements in the first row

• There are 5 teachers to be seated in the first row.

• The principal has a fixed seat, so only the other 4 teachers can be arranged.

• Number of ways to arrange the 4 teachers in the remaining 4 seats = 4! = 24.

Step 2: Arrangements in the second row

• There are 4 girls to be seated.

• Number of ways to arrange 4 girls in 4 seats = 4! = 24.

Step 3: Arrangements in the third row

• There are 6 boys to be seated.

• Number of ways to arrange 6 boys in 6 seats = 6! = 720.

Step 4: Total number of arrangements

Total arrangements = (arrangements in first row) × (arrangements in second row) × (arrangements in third row)

= 4! × 4! × 6!

= 24 × 24 × 720

= 24 × (24 × 720)

First calculate 24 × 720 = 17280

Then multiply by 24: 17280 × 24 = 414,720

When seating people around a round table, the number of distinct arrangements is given by:

(n - 1)!

because rotating the whole arrangement doesn't create a new seating order.

Here, n = 8 people.

Number of ways = (8 - 1)! = 7! = 5040

Eight different beads can be arranged in a circular form in (8-1)!=7! Ways. Since there is no distinction between the clockwise and anticlockwise arrangement, the required number of arrangements is 
7! /2 = 2520
 

Step 1: Number of ways to go from A to D via B and C

The total ways to go from A to D = (ways A→B) × (ways B→C) × (ways C→D)

= 3 × 5 × 2 = 30 ways.

Step 2: Number of ways to come back from D to A taking different roads than going

• For D → C, roads are same as C → D, total 2 roads. He must take a different one from the one he took going.

  • So if he took 1 road going, he has 1 different road coming back.

• For C → B, roads same as B → C (5 roads).

  • Similarly, he must choose a different road on return → 4 choices (5 - 1).

• For B → A, roads same as A → B (3 roads).

  • He must choose a different road on return → 2 choices (3 - 1).

Step 3: Total ways to return taking different roads

For each way going, number of ways coming back =

= (number of ways to take different roads on D→C) × (C→B) × (B→A)

= 1 × 4 × 2 = 8 ways.

Step 4: Total possible ways for the complete journey

= (ways going) × (ways returning with different roads)

= 30 × 8 = 240

Here is the extracted text without using LaTeX or KaTeX:

Step 1:
Consider the arrangements as circular permutations.

• Since the beads of the same color must come together, treat each color as a single unit.

• There are 9 different colors, so there are 9 units to arrange in a circle.

• The number of circular permutations of n distinct objects is (n - 1) !.

• In this case, the number of circular permutations is (9 - 1) ! = 8 !

Step 2: Account for the necklace being able to be flipped.

• A necklace can be flipped, so each arrangement in the circular permutation is counted twice.

• To correct for this, divide the number of circular permutations by 2.

• The number of arrangements is (8 !) /  2.

Step 3:Calculate the result.

• Calculate 8 ! = 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 = 40320.

• Divide by 2: 40320 /  2 = 20160.

Step 1: Understand the board

• The chessboard has 64 squares: 32 white squares and 32 black squares.

• Each row contains exactly 4 white squares and 4 black squares, alternating.

• So, each row has 4 white squares and 4 black squares.

Step 2: Pick one white square

• Suppose the white square is in a particular row (say row r).

• Since the white square is fixed, the row r is fixed.

Step 3: Black squares NOT in the same row

• Total black squares on the board = 32.

• Black squares in the same row r = 4 (since each row has 4 black squares).

• Black squares not in row r = 32 - 4 = 28.

Step 4: Conclusion

For any chosen white square (in row r), the number of black squares that are not in the same row = 28.

The correct answer is 2952 .

the solution is simple !!!

At least 5 from 8 means we have 5,6,7, and 8.

for necklace or circle we take (n-1)!, then

4!+5!+6!+7!= 5904.

For necklace we have pairs (mirror), so we divide on 2.

5904/2= 2952

To determine how many four-digit even numbers can be created from the prime numbers less than 10, we first identify the relevant prime numbers:

• 2
• 3
• 5
• 7

Since we want to form even numbers, the last digit must be 2. The process can be broken down as follows:

• Choose the last digit: 2 (fixed).
• Select the first three digits from the remaining primes: 3, 5, 7.

We can arrange these three digits in the first three positions:

• There are 3! (3 factorial) ways to arrange 3 digits.
• This equals 6 arrangements.

Thus, the total number of four-digit even numbers that can be formed is 6.