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Question

The number of ways that '6' rings can be worn on the 4 - fingers of one hand is

$4096$
$6^{4}$
$^{6} P_{4}$
6!

hint Hint:

We will first start by finding the way in which one ring can be worn in 4 fingers. Then we will do the same for 6 rings and then using the fundamental principle of counting we will find the total ways.

The correct answer is: $4096$

Complete step-by-step answer:
Now, we have been given that there are 6 rings of different types and we have to find the ways in which they can be worn in 4 fingers.

Now, we know that the number of options each ring has is 4, that is each ring has 4 fingers as their possible way as it can be worn in any one of 4 fingers.

Now, similarly the other rings will have four options as it has not been mentioned in the options that there has to be at least a ring in a finger. So, each ring has four options i.e. four fingers.
$N o w comma space w e space k n o w space t h a t space b y space t h e space f u n d a m e n t a l space p r i n c i p l e space o f space c o u n t i n g space t h e r e space c a n space b e space 4 cross times 4 cross times 4 cross times 4 cross times 4 cross times 4 space w a y s space o f space w e a r i n g space 6 space r i n g s S o comma space w e space h a v e space 4 to the power of 6 space equals space 4096 space w a y s space t o space w e a r space 6 space t y p e s space o f space r i n g s.$

It is important to note that we have used a basic fundamental principle of counting to find the total ways. Also, it is important to notice that each ring has 4 ways as it has not been given that each finger must have at least one ring. So, there can be 6 rings in a finger alone and remaining all the fingers empty.

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The number of ways that '6' rings can be worn on the 4 - fingers of one hand is

6!

We will first start by finding the way in which one ring can be worn in 4 fingers. Then we will do the same for 6 rings and then using the fundamental principle of counting we will find the total ways.

The correct answer is:

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$4096$
$6^{4}$
$^{6} P_{4}$
6!

The correct answer is: $4096$

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Tangents are drawn to the ellipse $fraction numerator x to the power of 2 end exponent over denominator 9 end fraction plus fraction numerator y to the power of 2 end exponent over denominator 5 end fraction equals 1$ at the ends of the latus rectum. The are of the quadrilateral so formed is

Tangents are drawn to the ellipse $fraction numerator x to the power of 2 end exponent over denominator 9 end fraction plus fraction numerator y to the power of 2 end exponent over denominator 5 end fraction equals 1$ at the ends of the latus rectum. The are of the quadrilateral so formed is

A tangent to the ellipse x² + 4y² = 4 meets the ellipse x² + 2y² = 6 at P and Q. The angle between the tangents at P and Q of the ellipse x² + 2y² = 6 is

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A cubic polynomial f(x) = ax³ + bx² + cx + d has a graph which touches the x-axis at 2, has another x-intercept at –1 and has y-intercept at –2 as shown. The value of, a + b + c + d equals

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The value of c in Lagrange’s theorem for the function f(x) = log sin x in the interval $open square brackets fraction numerator pi over denominator 6 end fraction comma fraction numerator 5 pi over denominator 6 end fraction close square brackets$ is -

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