Have a query? Ask a Tutor!!

Access to best educators and exclusive paper discussions

Can’t find what you’re looking for?

Our tutors are from top schools around the world, and they are waiting for you 24/7, anytime, anywhere.

Homework- One to One Solutions
Understand concepts to solve better
Get your doubts solved instantly

Maths-

General

Easy

Question

The number of 4 letter words that can be formed using the letters of the word EXPLAIN which begin with a vowel when repetitions are allowed is

1029
207
343
2401

Hint:

In the question we know that from the word EXPLAIN we have 3 vowels and to form the new 4 letter word which will start with vowel we will first choose the first letter from the 3 vowels only. Then we know that as the letters can be repeated so the next three letters can be chosen accordingly from all the 7 letters.

The correct answer is: 1029

In the word EXPLAIN, no. of vowels=3 no. of consonants =4
As the 4 letter word has to start with a vowel,
So, the number of ways of choosing the first letter = $blank cubed C subscript 1$ =3
Now for the next three letters we know that any letter from the 7 letters can be used as repetition is allowed.
Therefore, the number of 4 letter words that can be formed using the letters of the word EXPLAIN which begin with a vowel when repetitions are allowed = $3 cross times 7 cross times 7 cross times 7$ =1029

Related Questions to study

Which of the following statements is wrong?

Which of the following statements is wrong?

physics-General

The number of 3 digited numbers that can be formed using {1,2,3,4,5} any number of times is

The number of 3 digited numbers that can be formed using {1,2,3,4,5} any number of times =

5 cross times 5 cross times 5

The number of 3 digited numbers that can be formed using {1,2,3,4,5} any number of times is

The number of 3 digited numbers that can be formed using {1,2,3,4,5} any number of times =

5 cross times 5 cross times 5

If $blank to the power of 12 P subscript r equals to the power of 11 P subscript 6 plus 6 times to the power of 11 P subscript 5$ then r is :

blank to the power of 12 P subscript r equals to the power of 11 P subscript 6 plus 6 times to the power of 11 P subscript 5

rightwards double arrow fraction numerator 12 factorial over denominator open parentheses 12 minus r close parentheses factorial end fraction equals fraction numerator 11 factorial over denominator open parentheses 11 minus 6 close parentheses factorial end fraction plus 6. fraction numerator 11 factorial over denominator open parentheses 11 minus 5 close parentheses factorial end fraction rightwards double arrow fraction numerator 12 factorial over denominator open parentheses 12 minus r close parentheses factorial end fraction equals 11 factorial open square brackets fraction numerator 1 over denominator 5 factorial end fraction plus 6 fraction numerator 1 over denominator 6 factorial end fraction close square brackets rightwards double arrow fraction numerator 12 cross times 11 factorial over denominator open parentheses 12 minus r close parentheses factorial end fraction equals 11 factorial open square brackets fraction numerator 1 over denominator 5 factorial end fraction plus fraction numerator 6 over denominator 5 factorial cross times 6 end fraction close square brackets rightwards double arrow fraction numerator 12 over denominator open parentheses 12 minus r close parentheses factorial end fraction equals fraction numerator 2 over denominator 5 factorial end fraction rightwards double arrow fraction numerator 6 over denominator open parentheses 12 minus r close parentheses factorial end fraction equals fraction numerator 1 over denominator 5 factorial end fraction rightwards double arrow 6 cross times 5 factorial equals open parentheses 12 minus r close parentheses factorial rightwards double arrow 6 factorial equals open parentheses 12 minus r close parentheses factorial rightwards double arrow r equals 6

If $blank to the power of 12 P subscript r equals to the power of 11 P subscript 6 plus 6 times to the power of 11 P subscript 5$ then r is :

blank to the power of 12 P subscript r equals to the power of 11 P subscript 6 plus 6 times to the power of 11 P subscript 5

rightwards double arrow fraction numerator 12 factorial over denominator open parentheses 12 minus r close parentheses factorial end fraction equals fraction numerator 11 factorial over denominator open parentheses 11 minus 6 close parentheses factorial end fraction plus 6. fraction numerator 11 factorial over denominator open parentheses 11 minus 5 close parentheses factorial end fraction rightwards double arrow fraction numerator 12 factorial over denominator open parentheses 12 minus r close parentheses factorial end fraction equals 11 factorial open square brackets fraction numerator 1 over denominator 5 factorial end fraction plus 6 fraction numerator 1 over denominator 6 factorial end fraction close square brackets rightwards double arrow fraction numerator 12 cross times 11 factorial over denominator open parentheses 12 minus r close parentheses factorial end fraction equals 11 factorial open square brackets fraction numerator 1 over denominator 5 factorial end fraction plus fraction numerator 6 over denominator 5 factorial cross times 6 end fraction close square brackets rightwards double arrow fraction numerator 12 over denominator open parentheses 12 minus r close parentheses factorial end fraction equals fraction numerator 2 over denominator 5 factorial end fraction rightwards double arrow fraction numerator 6 over denominator open parentheses 12 minus r close parentheses factorial end fraction equals fraction numerator 1 over denominator 5 factorial end fraction rightwards double arrow 6 cross times 5 factorial equals open parentheses 12 minus r close parentheses factorial rightwards double arrow 6 factorial equals open parentheses 12 minus r close parentheses factorial rightwards double arrow r equals 6

parallel

Cyclic polymetaphosphates have the general formula of $open square brackets P subscript n end subscript O subscript 3 n end subscript close square brackets to the power of n minus end exponent$ Cyclic trimeric polymetaphosphate, $P subscript 3 O subscript 9 to the power of 3 minus end exponent$ is a compound heaving commercial importance. Which of the following is/are true regarding $P subscript 3 O subscript 9 to the power of 3 minus end exponent$ -

Cyclic polymetaphosphates have the general formula of $open square brackets P subscript n end subscript O subscript 3 n end subscript close square brackets to the power of n minus end exponent$ Cyclic trimeric polymetaphosphate, $P subscript 3 O subscript 9 to the power of 3 minus end exponent$ is a compound heaving commercial importance. Which of the following is/are true regarding $P subscript 3 O subscript 9 to the power of 3 minus end exponent$ -

chemistry-General

If $scriptbase P subscript 7 end scriptbase presuperscript n equals 42 times to the power of n P subscript 5$ then n is :

scriptbase P subscript 7 end scriptbase presuperscript n equals 42 times to the power of n P subscript 5

rightwards double arrow fraction numerator n factorial over denominator open parentheses n minus 7 close parentheses factorial end fraction equals 42. fraction numerator n factorial over denominator open parentheses n minus 5 close parentheses factorial end fraction rightwards double arrow fraction numerator open parentheses n minus 5 close parentheses factorial over denominator open parentheses n minus 7 close parentheses factorial end fraction equals 42 rightwards double arrow fraction numerator open parentheses n minus 5 close parentheses open parentheses n minus 6 close parentheses open parentheses n minus 7 close parentheses factorial over denominator open parentheses n minus 7 factorial close parentheses end fraction equals 42 rightwards double arrow open parentheses n minus 5 close parentheses open parentheses n minus 6 close parentheses equals 7 cross times 6 rightwards double arrow open parentheses 12 minus 5 close parentheses open parentheses 12 minus 6 close parentheses equals 7 cross times 6 rightwards double arrow n equals 12

If $scriptbase P subscript 7 end scriptbase presuperscript n equals 42 times to the power of n P subscript 5$ then n is :

scriptbase P subscript 7 end scriptbase presuperscript n equals 42 times to the power of n P subscript 5

rightwards double arrow fraction numerator n factorial over denominator open parentheses n minus 7 close parentheses factorial end fraction equals 42. fraction numerator n factorial over denominator open parentheses n minus 5 close parentheses factorial end fraction rightwards double arrow fraction numerator open parentheses n minus 5 close parentheses factorial over denominator open parentheses n minus 7 close parentheses factorial end fraction equals 42 rightwards double arrow fraction numerator open parentheses n minus 5 close parentheses open parentheses n minus 6 close parentheses open parentheses n minus 7 close parentheses factorial over denominator open parentheses n minus 7 factorial close parentheses end fraction equals 42 rightwards double arrow open parentheses n minus 5 close parentheses open parentheses n minus 6 close parentheses equals 7 cross times 6 rightwards double arrow open parentheses 12 minus 5 close parentheses open parentheses 12 minus 6 close parentheses equals 7 cross times 6 rightwards double arrow n equals 12

STATEMENT EXPLANATION Although $P F subscript 5 end subscript comma P C l subscript 5 end subscript$ andPhosphorus has $P B r subscript 5 end subscript$ are known, thelower pentahalides of nitrogenelectronegativity have not been observedthan nitrogen.electronegativity

STATEMENT EXPLANATION Although $P F subscript 5 end subscript comma P C l subscript 5 end subscript$ andPhosphorus has $P B r subscript 5 end subscript$ are known, thelower pentahalides of nitrogenelectronegativity have not been observedthan nitrogen.electronegativity

chemistry-General

parallel

Write the missing product in the following reaction– $4 H N O subscript 3 plus P subscript 4 O subscript 10 rightwards arrow 4 H P O subscript 3$

Write the missing product in the following reaction– $4 H N O subscript 3 plus P subscript 4 O subscript 10 rightwards arrow 4 H P O subscript 3$

chemistry-General

In phosphorus estimation, the compound Mg NH₄PO₄ is converted on ignition to–

In phosphorus estimation, the compound Mg NH₄PO₄ is converted on ignition to–

chemistry-General

If $scriptbase P subscript r end scriptbase presuperscript 8 equals to the power of 8 P subscript left parenthesis r plus 1 right parenthesis$ then r is :

G i v e n space to the power of 8 P subscript r equals blank to the power of 8 P subscript open parentheses r plus 1 close parentheses end subscript rightwards double arrow fraction numerator 8 factorial over denominator open parentheses 8 minus r close parentheses factorial end fraction equals fraction numerator 8 factorial over denominator open parentheses 8 minus r minus 1 close parentheses factorial end fraction rightwards double arrow fraction numerator 1 over denominator open parentheses 8 minus r close parentheses factorial end fraction equals fraction numerator 1 over denominator open parentheses 7 minus r close parentheses factorial end fraction rightwards double arrow open parentheses 7 minus r close parentheses factorial equals open parentheses 8 minus r close parentheses factorial rightwards double arrow open parentheses 7 minus 7 close parentheses factorial equals open parentheses 8 minus 7 close parentheses factorial space space space space space open square brackets 0 factorial equals 1 space a n d space 1 factorial equals 1 close square brackets rightwards double arrow r equals 7

If $scriptbase P subscript r end scriptbase presuperscript 8 equals to the power of 8 P subscript left parenthesis r plus 1 right parenthesis$ then r is :

G i v e n space to the power of 8 P subscript r equals blank to the power of 8 P subscript open parentheses r plus 1 close parentheses end subscript rightwards double arrow fraction numerator 8 factorial over denominator open parentheses 8 minus r close parentheses factorial end fraction equals fraction numerator 8 factorial over denominator open parentheses 8 minus r minus 1 close parentheses factorial end fraction rightwards double arrow fraction numerator 1 over denominator open parentheses 8 minus r close parentheses factorial end fraction equals fraction numerator 1 over denominator open parentheses 7 minus r close parentheses factorial end fraction rightwards double arrow open parentheses 7 minus r close parentheses factorial equals open parentheses 8 minus r close parentheses factorial rightwards double arrow open parentheses 7 minus 7 close parentheses factorial equals open parentheses 8 minus 7 close parentheses factorial space space space space space open square brackets 0 factorial equals 1 space a n d space 1 factorial equals 1 close square brackets rightwards double arrow r equals 7

parallel

Solid $P C I subscript 5$ exists as–

Solid $P C I subscript 5$ exists as–

chemistry-General

The coefficient of $x to the power of n$ in $fraction numerator 1 over denominator left parenthesis 1 minus x right parenthesis left parenthesis 1 minus 2 x right parenthesis left parenthesis 1 minus 3 x right parenthesis end fraction$ is

The coefficient of $x to the power of n$ in $fraction numerator 1 over denominator left parenthesis 1 minus x right parenthesis left parenthesis 1 minus 2 x right parenthesis left parenthesis 1 minus 3 x right parenthesis end fraction$ is

Statement I Two particles moving in the same direction do not lose all their energy in a completely inelastic collision.
Statement II Principle of conservation of momentum holds true for all kinds of collisions.

Statement I Two particles moving in the same direction do not lose all their energy in a completely inelastic collision.
Statement II Principle of conservation of momentum holds true for all kinds of collisions.

physics-General

parallel

The coefficient of $x to the power of n$ in $fraction numerator x minus 4 over denominator x squared minus 5 x plus 6 end fraction$ is

The coefficient of $x to the power of n$ in $fraction numerator x minus 4 over denominator x squared minus 5 x plus 6 end fraction$ is

(Me)₂SiCl₂ on hydrolysis will produce -

left parenthesis M e right parenthesis subscript 2 end subscript S i C l subscript 2 end subscript stack ⟶ with H subscript 2 end subscript O on top open square brackets negative O minus left parenthesis M e right parenthesis subscript 2 end subscript S i minus O minus close square brackets subscript n end subscript

(Me)₂SiCl₂ on hydrolysis will produce -

chemistry-General

left parenthesis M e right parenthesis subscript 2 end subscript S i C l subscript 2 end subscript stack ⟶ with H subscript 2 end subscript O on top open square brackets negative O minus left parenthesis M e right parenthesis subscript 2 end subscript S i minus O minus close square brackets subscript n end subscript

The coefficient of $x to the power of n$ in $fraction numerator left parenthesis 1 plus x right parenthesis left parenthesis 1 plus 2 x right parenthesis left parenthesis 1 plus 3 x right parenthesis over denominator left parenthesis 1 minus x right parenthesis left parenthesis 1 minus 2 x right parenthesis left parenthesis 1 minus 3 x right parenthesis end fraction$ is

The coefficient of $x to the power of n$ in $fraction numerator left parenthesis 1 plus x right parenthesis left parenthesis 1 plus 2 x right parenthesis left parenthesis 1 plus 3 x right parenthesis over denominator left parenthesis 1 minus x right parenthesis left parenthesis 1 minus 2 x right parenthesis left parenthesis 1 minus 3 x right parenthesis end fraction$ is

parallel

card img

With Turito Academy.

card img

With Turito Foundation.

card img

Get an Expert Advice From Turito.

Turito Academy

card img

With Turito Academy.

Test Prep

card img

With Turito Foundation.