How many different nine digit numbers can be formed from the number 2,2,3,3,5,5,8,8,8 by rearranging its digits so that the odd digits occupy even position ?

The sum of all the numbers that can be formed with the digits 2, 3, 4, 5 taken all at a time is (repetition is not allowed) :

Alternatively, we can use the formula for the sum of numbers as
$left parenthesis n minus 1 right parenthesis factorial space cross times space left parenthesis S u m space o f space d i g i t s right parenthesis space cross times left parenthesis 11111...... space n t i m e s right parenthesis$
We can also solve this problem by writing all the possible numbers and finding the sum of them which will be time taking and make us confused. We should know that the value of the digits is determined by the place where they were present. We should check whether there is zero in the given digits and whether there are any repetitions present in the numbers. Similarly, we can expect problems to find the sum of numbers formed by these digits with repetition allowed.

The sum of all the numbers that can be formed with the digits 2, 3, 4, 5 taken all at a time is (repetition is not allowed) :

Total number of divisors of 480, that are of the form 4n + 2, n $greater or equal than$ 0, is equal to :

We can also solve this question by writing 4n + 2 = 2(2n + 1) where 2n + 1 is always an odd number. So, when all odd divisors will be multiplied by 2, we will get the divisors that we require. Hence, we can say a number of divisors of 4n + 2 form is the same as the number of odd divisors for 480.

Total number of divisors of 480, that are of the form 4n + 2, n $greater or equal than$ 0, is equal to :

If ⁹P₅ + 5 ⁹P₄ = $scriptbase P subscript r end scriptbase presuperscript 10$ , then r =

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If ⁹P₅ + 5 ⁹P₄ = $scriptbase P subscript r end scriptbase presuperscript 10$ , then r =

maths-

Assertion (A) :If $fraction numerator 5 x plus 1 over denominator left parenthesis x plus 2 right parenthesis left parenthesis x minus 1 right parenthesis end fraction equals fraction numerator A over denominator x plus 2 end fraction plus fraction numerator B over denominator x minus 1 end fraction$ , then $A equals 3 comma B equals 2$
Reason (R) : $fraction numerator P x plus q over denominator left parenthesis x minus a right parenthesis left parenthesis x minus b right parenthesis end fraction equals fraction numerator P a plus q over denominator left parenthesis x minus a right parenthesis left parenthesis a minus b right parenthesis end fraction plus fraction numerator P b plus q over denominator left parenthesis x minus b right parenthesis left parenthesis b minus a right parenthesis end fraction$

maths-General

physics-

A particle is released from a height. At certain height its kinetic energy is three times its potential energy. The height and speed of the particle at that instant are respectively

physics-General

If x is real, then maximum value of $fraction numerator 3 x to the power of 2 end exponent plus 9 x plus 17 over denominator 3 x to the power of 2 end exponent plus 9 x plus 7 end fraction$ is

The value of 'c' of Lagrange's mean value theorem for $f space left parenthesis x right parenthesis equals x squared minus 3 x minus 2 text for end text x element of left square bracket negative 1 , 2 right square bracket$ is

The value of 'c' of Rolle's mean value theorem for $f space left parenthesis x right parenthesis equals vertical line x vertical line equals i n space left square bracket negative 1 , 1 right square bracket$ is

The value of 'c' of Rolle's theorem for $f space left parenthesis x right parenthesis equals l o g space open parentheses x squared plus 2 close parentheses$ – $l o g subscript 3 end subscript$ on [–1, 1] is

For $f space left parenthesis x right parenthesis equals 4 minus left parenthesis 6 minus x right parenthesis to the power of 2 divided by 3 end exponent$ in [5, 7]

The value of 'c' in Lagrange's mean value theorem for $f space left parenthesis x right parenthesis equals x cubed minus 2 x squared minus x plus 4$ in [0, 1] is

The value of 'c' in Lagrange's mean value theorem for $f space left parenthesis x right parenthesis equals x space left parenthesis x minus 2 right parenthesis squared$ in [0, 2] is

The equation $r equals a c o s space theta plus b s i n space theta$ represents

The polar equation of the circle whose end points of the diameter are $open parentheses square root of 2 comma fraction numerator pi over denominator 4 end fraction close parentheses$ and $open parentheses square root of 2 comma fraction numerator 3 pi over denominator 4 end fraction close parentheses$ is