How many seven-letter passwords can be made from the letters of the En

Question

How many seven-letter passwords can be made from the letters of the English alphabet, so that letter "A" is in the middle and no repetition is allowed?

A. 167,765,600
B. 127,512,000
C. 6,375,600
D. 5,040
E. 720

Amogh223 · Answer

Ans is d 
2 cases when aa is included and when single a is included 
720+ 4320= 5040

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akadiyan · Answer

How many seven-letter passwords can be made from the letters of the English alphabet, so that letter "A" is in the middle and no repetition is allowed?

fall2021 
Thanks for pointing out the error

when seven-letter passwords can be made from the letters of english alphabets, the logic remains the same as given below
I get 25*25*24*23*22*21*20.

Can you explain the logic behind your solution or what am missing?

------------------------------------------------------------------------------------------------------------------------------------------------------
Note: The question did not ask for seven-letter password form the word 'Alphabet', hence this answer is incorrect.

Digit 1: 6 options (Cannot have A , since there are 2 A's in alphabet, there are only 6 possible options)
Digit 2: 7 options (Can have both the A's and all other digits except the one used in digit 1)
Digit 3: 5 options
Digit 4: 4 options
Digit 3: 3 options
Digit 2: 2 options
Digit 1: 1 option

Total number of possible combinations = 6*7*5*4*3*2*1 = 5040

Ans:   D

fall2021 · Answer

The question says English alphabet, which is all 26 letters from A to Z. It doesn't talk about the word 'alphabet'

The answer I got is B

25C3*22C3*3!*3! = 127,512,000

Or simply 25P6 which is the same number. 'A' being in the middle is not affecting anything as each letter in the password is unique.

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fall2021 · Answer

Hi, you are almost there, my answer is 25P6 which is same as 25*24*23*22*21*20. This will give option B.

You wrote an extra 25. We get to choose only 6 letters because A is already chosen.

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CrackverbalGMAT · Answer

Use the slot/boxes method..There are 26 Alphabets, of which A occupies the 4th position.

The first position can be occupied in 25 ways.

The second position can be occupied in 24 ways.

The third position can be occupied in 23 ways.

The fourth position can be occupied in 1 way.

The fifth position can be occupied in 22 ways.

The sixth position can be occupied in 21 ways.

The seventh position can be occupied in 20 ways.

Total ways = 25 * 24 * 23 * 1 * 22 * 21 * 20 = 127,512,000

Option B

Arun Kumar

akadiyan · Answer

Ah got it, i missed a minor but important detail. Thanks for the explanation

Aderonke01 · Answer

The middle number would be A. List 7 lines. Put A in the middle, being the first limitation can be used in 1 way. We have 25 other possibilities. From 26 alphabet.

— — — — — — —
25 * 24 * 23 * 1 * 22*21*20=127, 512, 000. IMO B

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rito54 · Answer

Hello, how do we compute quickly this calculation, it is quite long to do 25*24*23*1*22*21*20 no ? Thanks

shamid19 · Answer

Is there an easier way to calculate the solution other than manually do 25*24*23*22*21*20?

I tried 20*20*20*20*20*20 or 2^6 but it results in something with 64.

Any suggestions?

sarthak1701 · Answer

Try to first multiply numbers that DO NOT end in 0 (or 5 in any other case)

Here we have 25*24*23*1*22*21*20 possibilities (multiplied by 1 because A stays in the same position)

Split this into 2, 3 numbers each,

we get 25*352*462*20

this results in 8750*9240

The last 2 digits of these numbers are 40 and 50 which means the product should end in 2000 and there's only one option ending in 2000.

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How many seven-letter passwords can be made from the letters of the En

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How many seven-letter passwords can be made from the letters of the En

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