Author: admin

The product of all solutions of the equation \[ e^{5(\log_e{x})^2 + 3} = x^8, \quad x > 0 \] is: \( e^{\frac{8}{5}} \)\( e^{\frac{6}{5}} \)\( e^{2} \)\( e \)

Definition 1. (&y=f(x)&x=a,x=b&(x-ax” is ” )/b&” Area ” =∫_a^t  t(x)dx.)

#. Area bounded by curve and axis: Definition 1. Suppose that $(X,\mathcal M)$ and $(Y,\mathcal N)$ are measurable spaces, and $f:X\to Y$ is a map. We call $f$ is measurable if for every $B\in\mathcal N$ the set $f^{-1}(B)$ is in $\mathcal M$.

Given numbers – 1000000 Number of digits = 7 The numbers have to be greater than 1000000, so they can begin either with 1, 2 or 4. When 1 is fixed at the extreme left position, the remaining digits to be rearranged will be 0, 2, 2, 2, 4, 4, in which there are 3, 2s and 2, 4s. Thus, the number of numbers beginning with 1 = 6!/(3! 2!) = (6 × 5 × 4 × 3!)/(3! × 2) … Continue reading “How many numbers greater than 1000000 can be formed using the digits 1, 2, 0, 2, 4, 2, 4?”

There are 4 Ace cards in a deck of 52 cards. According to the given, we need to select 1 Ace card out of the 4 Ace cards ∴ The number of ways to select 1 Ace from 4 Ace cards is 4C1 ⇒ More 4 cards are to be selected now from 48 cards (52 cards – 4 Ace cards) ∴ The number of ways to select 4 cards from 48 cards is 48C4 Number of 5 card combinations out of … Continue reading “Determine the number of 5 card combinations out of a deck of 52 cards, if there is exactly one ace in each combination.”

Men = 2 Women = 3 A committee of 3 persons to be constituted. Here, the order does not matter. Therefore, we need to count combinations. There will be as many committees as combinations of 5 different persons taken 3 at a time. Hence, the required number of ways = 5C3 = 5!/(3! 2!) = (5 × 4 × 3!)/(3! × 2) = 10 Committees with 1 man and 2 women: 1 man can be selected from 2 men in 2C1 ways. 2 … Continue reading “A committee of 3 persons is to be constituted from a group of 2 men and 3 women. In how many ways can this be done? How many of these committees would consist of 1 man and 2 women?”

Total number of families = 87 Number of families with at most 2 children = 52 Remaining families = 87 – 52 = 35 Also, for the rural development programme, 20 families are to be chosen for assistance, of which at least 18 families must have at most 2 children. Thus, the following are the number of possible choices: 52C18 × 35C2 (18 families having at most 2 children and 2 selected from other types of families) 52C19 × 35C1 (19 families having at most 2 … Continue reading “In a small village, there are 87 families, of which 52 families have at most 2 children. In a rural development programme, 20 families are to be chosen for assistance, of which at least 18 families must have at most 2 children. In how many ways can the choice be made?”

Given word – MISSISSIPPI M – 1 I – 4 S – 4 P – 2 Number of permutations = 11!/(4! 4! 2!) = (11 × 10 × 9 × 8 × 7 × 6 × 5 × 4!)/ (4! × 24 × 2) = 34650 We take that 4 I’s come together, and they are treated as 1 letter, ∴ Total number of letters=11 – 4 + 1 = 8 ⇒ Number of permutations = 8!/(4! 2!) = (8 … Continue reading “In how many of the distinct permutations of the letters in MISSISSIPPI do the four Is not come together?”

Let ABCDE be a five-digit number. Given that the first two digits of each number are 6 and 7. So, the number is 67CDE. As repetition is not allowed and 6 and 7 are already taken, the digits available for place C are 0, 1, 2, 3, 4, 5, 8, 9, i.e. eight possible digits. Suppose one of them is taken at C, now the digits possible at place D is 7. Similarly, at E, the possible digit is 6. … Continue reading “How many 5-digit telephone numbers can be constructed using the digits 0 to 9, if each number starts with 67 and no digit appears more than once?”

Numbers between 99 and 1000 are all three-digit numbers. Total number of 3 digit numbers having at least one of their digits as 7 = (Total numbers of three-digit numbers) – (Total number of 3 digit numbers in which 7 does not appear at all) = (9 × 10 × 10) – (8 × 9 × 9) = 900 – 648 = 252