Maths-

General

Easy

Question

Number of permutations that can be made using all the letters of the word MATRIX is

120
240
480
720

hint Hint:

Basically Permutation is an arrangement of objects in a particular way or order. While dealing with permutation one should concern about the selection as well as arrangement.
In Short, ordering is very much essential in permutations. In other words, the permutation is considered as an ordered combination.

The correct answer is: 720

The word MATRIX is a 6 letter word
Thus, number of permutations that can be made using all the letters of the word MATRIX is 6! = 720

The number of different signals that can be made by 5 flags from 8 flags of different colours is :

Each of the different arrangements which can be made by taking some or all of a number of things at a time is called permutation.
$space T h e space n u m b e r space o f space p e r m u t a t i o n space w i t h o u t space a n y space r e p e t i t i o n space s t a t e s space t h a t space a r r a n g i n g space n space space o b j e c t s space t a k e n space r space space a t space a space t i m e space i s space e q u i v a l e n t space t o space f i l l i n g space r space space p l a c e s space o u t space o f space n space space t h i n g s space equals space n left parenthesis n minus 1 right parenthesis left parenthesis n minus 2 right parenthesis... left parenthesis n minus top enclose r minus 1 end enclose right parenthesis space space space w a y s equals fraction numerator n factorial over denominator left parenthesis n minus r right parenthesis factorial end fraction space equals space scriptbase P subscript r end scriptbase presuperscript n$

$W e space c a n space a l s o space s a y space t h a t comma space space scriptbase P subscript r end scriptbase presuperscript n equals space n left parenthesis n minus 1 right parenthesis left parenthesis n minus 2 right parenthesis... left parenthesis n minus r plus 1 right parenthesis space space comma space t h u s comma scriptbase P subscript 5 end scriptbase presuperscript 8 equals space 8 cross times 7 cross times 6 cross times 5 cross times 4 space equals space 6720 space.$

Thus we can say that the permutation concerns both the selection and the arrangement of the selected things in all possible ways.

Have a query? Ask a Tutor!!

Can’t find what you’re looking for?

Number of permutations that can be made using all the letters of the word MATRIX is

120 240 480 720

The correct answer is: 720

The word MATRIX is a 6 letter wordThus, number of permutations that can be made using all the letters of the word MATRIX is 6! = 720

Related Questions to study

The number of different signals that can be made by 5 flags from 8 flags of different colours is :

The number of different signals that can be made by 5 flags from 8 flags of different colours is :

If then A-B=

If then A-B=

, then

, then

If then K=

If then K=

If the remainders of the polynomial when divided by and are 7,3 then the remainder of when devided by is

If the remainders of the polynomial when divided by and are 7,3 then the remainder of when devided by is

If then

If then

The partial fractions of are

The partial fractions of are

The partial fractions of are

The partial fractions of are

then

then

The partial fractions of are

The partial fractions of are

The partial fractions of are

The partial fractions of are

If then

If then

Let a,b,c such that then ,?

Let a,b,c such that then ,?

If then B=

If then B=

If the remainders of the polynomial f(x) when divided by are 2,5 then the remainder of when divided by is

If the remainders of the polynomial f(x) when divided by are 2,5 then the remainder of when divided by is

With Turito Academy.

With Turito Foundation.

Get an Expert Advice From Turito.

Turito Academy

With Turito Academy.

Test Prep

With Turito Foundation.

Courses

Quick Links

120
240
480
720

The word MATRIX is a 6 letter word
Thus, number of permutations that can be made using all the letters of the word MATRIX is 6! = 720

If the remainders of the polynomial $f left parenthesis x right parenthesis$ when divided by $x plus 1$ and $x minus 1$ are 7,3 then the remainder of $f left parenthesis x right parenthesis$ when devided by $x to the power of 2 end exponent minus 1$ is

If the remainders of the polynomial $f left parenthesis x right parenthesis$ when divided by $x plus 1$ and $x minus 1$ are 7,3 then the remainder of $f left parenthesis x right parenthesis$ when devided by $x to the power of 2 end exponent minus 1$ is

The partial fractions of $fraction numerator left parenthesis x plus 1 right parenthesis to the power of 2 end exponent over denominator x open parentheses x to the power of 2 end exponent plus 1 close parentheses end fraction$ are

The partial fractions of $fraction numerator left parenthesis x plus 1 right parenthesis to the power of 2 end exponent over denominator x open parentheses x to the power of 2 end exponent plus 1 close parentheses end fraction$ are

The partial fractions of $fraction numerator 1 over denominator x to the power of 3 end exponent left parenthesis x plus 2 right parenthesis end fraction$ are

The partial fractions of $fraction numerator 1 over denominator x to the power of 3 end exponent left parenthesis x plus 2 right parenthesis end fraction$ are

The partial fractions of $fraction numerator 6 x to the power of 4 end exponent plus 5 x to the power of 3 end exponent plus x to the power of 2 end exponent plus 5 x plus 2 over denominator 1 plus 5 x plus 6 x to the power of 2 end exponent end fraction$ are

The partial fractions of $fraction numerator 6 x to the power of 4 end exponent plus 5 x to the power of 3 end exponent plus x to the power of 2 end exponent plus 5 x plus 2 over denominator 1 plus 5 x plus 6 x to the power of 2 end exponent end fraction$ are

If the remainders of the polynomial f(x) when divided by $x minus 1 comma x minus 2$ are 2,5 then the remainder of $f left parenthesis x right parenthesis$ when divided by $left parenthesis x minus 1 right parenthesis left parenthesis x minus 2 right parenthesis$ is

If the remainders of the polynomial f(x) when divided by $x minus 1 comma x minus 2$ are 2,5 then the remainder of $f left parenthesis x right parenthesis$ when divided by $left parenthesis x minus 1 right parenthesis left parenthesis x minus 2 right parenthesis$ is