Permutation And Combination - JEE/KEAM

A polygon with 65 diagonals must have 14 sides. To find this, we can use the formula for the number of diagonals in a polygon, which is n(n-3)/2, where n is the number of sides. By substituting the given number of diagonals (65) into the formula, we can solve for n. Simplifying the equation, we get n^2 - 3n - 130 = 0. Factoring this equation, we find that (n-14)(n+9) = 0. Therefore, the possible values for n are -9 and 14. Since the number of sides cannot be negative, the polygon must have 14 sides.

When the letters of the word 'KUBER' are arranged in all possible orders and arranged as in a dictionary, the word 'KUBER' will have a rank of 67.

The word CALCUTTA has 8 letters, including 2 C's and 2 T's. To find the number of arrangements, we use the formula for permutations of a word with repeated letters, which is n! / (r1! * r2! * ... * rk!), where n is the total number of letters and r1, r2, ..., rk are the frequencies of each repeated letter. In this case, n = 8, r1 = 2 (for C), and r2 = 2 (for T). Plugging these values into the formula, we get 8! / (2! * 2!) = 5040. Therefore, the correct answer is 5040.

The total number of 4-digit numbers that can be formed using the digits 1, 2, 3, 4, 5, and 6 without repetition is equal to the total number of ways to arrange these digits in a 4-digit number. Since the first digit must be greater than 3, there are 3 options (4, 5, or 6) for the first digit. After choosing the first digit, there are 5 options remaining for the second digit, 4 options for the third digit, and 3 options for the fourth digit. Therefore, the total number of arrangements is 3 * 5 * 4 * 3 = 240.

The number of triangles that can be formed by joining the angular points of a decagon can be calculated using the formula n(n-1)(n-2)/6, where n is the number of angular points. In this case, the decagon has 10 angular points, so substituting n=10 into the formula gives us 10(10-1)(10-2)/6 = 10*9*8/6 = 120. Therefore, the correct answer is 120.

To form an even number with 3 different digits, the last digit must be 2, 4, 6, or 8. There are 4 choices for the last digit. For the first digit, we have 8 choices (any digit except 0 and the one already chosen for the last digit). For the second digit, we have 7 choices (any digit except the two already chosen). So, the total number of even numbers that can be formed is 4 x 8 x 7 = 224.

To form a four-digit number using the digits 0, 1, 2, and 3 without repeating any digit, we have the following options for the first digit:
For the thousands place: We have 3 options (1, 2, or 3).
For the hundreds place: We have 3 options (0, 1, or 2), excluding the digit used for the thousands place.
For the tens place: We have 2 options (remaining digits after choosing the thousands and hundreds place).
For the units place: We have 1 option (the remaining digit after choosing the thousands, hundreds, and tens place).
Therefore, the total number of four-digit numbers without repeated digits is 3 × 3 × 2 × 1 = 18.
Now, to find the number of even numbers among these, we consider the units place. Since the units place can only be filled with 0 or 2 to make the number even, there are 2 options for the units place.
Therefore, the number of even four-digit numbers without repeated digits is 3 × 3 × 2 × 1 × 2 = 36.
However, we need to consider that the number 0000 (which is not typically considered a four-digit number) is included in this count. So, we subtract 1 from the total.
Therefore, the correct answer is 36 - 1 = 35.
So, none of the given options is correct.

There are five examination papers, and we need to arrange them in such a way that the physics and chemistry papers never come together. One approach is to consider the physics and chemistry papers as a single unit. This means we have four units to arrange: the combined physics and chemistry unit, along with the remaining three papers. The four units can be arranged in 4! = 24 ways. Within the combined unit, the physics and chemistry papers can be arranged in 2! = 2 ways. Therefore, the total number of arrangements is 24 * 2 = 48. However, we need to consider that the physics and chemistry papers can also be arranged in the opposite order. So, the final answer is 48 * 2 = 72.

The total number of flags with three horizontal strips, in order that can be formed using 2 identical red, 2 identical green and 2 identical white strips, is equal to 4! (4 factorial). This is because there are 4 positions for the strips and each position can be filled with any of the 4 strips (2 red, 2 green, and 2 white). Therefore, the total number of possible arrangements is 4 x 3 x 2 x 1 = 24, which is equal to 4!.

The number of 5 digit telephone numbers having at least one of their digits repeated can be calculated using the principle of inclusion-exclusion. There are a total of 100,000 5-digit numbers possible. To find the numbers without any repeated digits, we have 10 choices for the first digit, 9 choices for the second digit (excluding the first), 8 choices for the third digit, 7 choices for the fourth digit, and 6 choices for the fifth digit. Therefore, the number of numbers without any repeated digits is 10 * 9 * 8 * 7 * 6 = 30,240. Subtracting this from the total number of 5-digit numbers gives us the answer of 69,760.

The number of all the odd divisors of 3600 can be found by prime factorizing 3600, which is 2^4 * 3^2 * 5^2. Since odd divisors can only have odd prime factors, we can choose any combination of the odd prime factors (3 and 5) and their powers. For each prime factor, we have (power + 1) choices. Therefore, for 3 we have (2+1) = 3 choices, and for 5 we also have (2+1) = 3 choices. Multiplying these choices together, we get 3 * 3 = 9, which is the number of all the odd divisors of 3600.

Each prize can be given to any of the four students, so there are 4 choices for each prize. Since there are 5 prizes, the total number of ways to distribute the prizes is 4^5 = 1024.

The number of ways in which the number of tails is equal to the number of heads can be calculated by finding the number of ways to arrange equal numbers of tails and heads in a row. Since there are 6 coins, there are 6 possible positions for the first tail/head, 5 positions for the second, 4 for the third, 3 for the fourth, 2 for the fifth, and 1 for the last. This gives us a total of 6! (6 factorial) ways to arrange the tails and heads. However, since the coins are identical, we need to divide this by the number of ways to arrange the identical coins, which is 3! (3 factorial). Therefore, the number of ways is 6! / (3!) = 20.

When rolling four dice, there are a total of 6^4 = 1296 possible outcomes. To find the number of outcomes in which at least one die shows 2, we can calculate the total number of outcomes in which no die shows 2 and subtract it from the total number of possible outcomes. Since there are 5 numbers on each die that are not 2, the number of outcomes in which no die shows 2 is 5^4 = 625. Therefore, the number of outcomes in which at least one die shows 2 is 1296 - 625 = 671.