How many 6 digit even numbers can be formed from the digits 1,2 3 4 5 SO 7 the digits should not repeat and second last digit is even?

The number of six-digit numbers that can be formed from the digits 1, 2, 3, 4, 5, 6 and 7 so that digits do not repeat, and the terminal digits are even is
A. 144
B. 72
C. 288
D. 720

Answer

Verified

Hint:Let us first fill the terminal digits of the number with even digits. After that fill all other places in a number. then find all the possible combinations and get the solution.

Complete step-by-step answer:
As we know that terminal digits in a number are only first and last digit.
So, let us fill the first and last place of the number with an even digit.
And we know that any digit is even if it is exactly divided by 2.
So, out of given digits there are only 3 even digits and that were {2, 4, 6}.
So, there are 3 possible digits that can be placed in the first place.
As we know that digits in a number are not repeated.
So, there will be 2 possible digits left for last place.
Now, we are left with 5 digits and 4 places. As we know that these four places can be filled with any number even or odd. But the digits cannot be repeated.
So, possible digits for second place will be 5.
Possible digits left for third place will be 4.
Possible digits left for fourth place will be 3.
And possible digits left for fifth place will be 2.
So, total number of numbers that can be formed with the digits {1, 2, 3, 4, 5, 6, 7} with no digits repeated and terminal digits as even will be 3$\times$5$\times$4$\times$3$\times$2$\times$2 = 720.
So, total possible numbers will be 720.
Hence, the correct option will be D.

Note: Whenever we come up with this type of problem then first, we will fill first and last place of the number with even digits and then other places. After placing a digit at any place, we subtract the number of digits left by 1. Because it is given that digits in a number cannot be repeated. And this will be the easiest and efficient way to find the solution of the problem.

Table of Contents

• The number of six-digit numbers that can be formed from the digits 1, 2, 3, 4, 5, 6 and 7 so that digits do not repeat, and the terminal digits are even isA. 144B. 72C. 288D. 720
• Terminal digits mean the first digit and the last digit. Here, it is given that the terminal digits are even.So, The first position can be filled in 3 ways i.e, by 2,4 and 6Since the repetition is not allowed and the terminal digits need to be even, and we have used one even digit at the first position, So now we are left with '2‘ even digits, So, The last place can be filled in ’2' ways.Now, we have left with '5; digits and '4' places. As we know that these four places can be filled with any number even or odd. But the digits cannot be repeated.So, the possible digits for second place will be 5Possible digits left for third place will be 4Possible digits left for fourth place will be 3And possible digits left for the fifth place will be 2So, the total number of numbers that can be formed with the digits {1,2,3, 4,5,6,7} with no digits repeated and terminal digits as even=3×5×4×3×2×2=720So, the total possible numbers will be 720.
• The number of six-digit numbers that can be formed from the digits 1, 2, 3, 4, 5, 6 and 7 so that digits do not repeat, and the terminal digits are even isA. 144B. 72C. 288D. 720
• How many 6 digit even numbers can be formed from digits 1 2 3 4 5 7 and that the digit should not repeat and second last digit is even?
• How many 5 digit even numbers can be formed with 1 2 3 4 5 if the digits do not repeat?
• How many 6 digit even numbers can be made from the digits 2146 3 and 5?
• How many 6 digit even numbers can be formed from digits 1 to 7 so that the digits should not repeat and the second last digit is even?
• How many 6 digit even numbers can be formed from digits 1 to 7 so that the digits should not repeat and the second last digit is even?
• How many 6 digit even numbers can be formed?
• How many 6 digit numbers can be formed from 1 to 7 which are divisible by 5?
• How many six digit numbers can be formed using the digits 1 to 6 without repetition such that the number is divisible by the digit at its units place?

The number of six-digit numbers that can be formed from the digits 1, 2, 3, 4, 5, 6 and 7 so that digits do not repeat, and the terminal digits are even isA. 144B. 72C. 288D. 720

Answer

Verified

Hint:Let us first fill the terminal digits of the number with even digits. After that fill all other places in a number. then find all the possible combinations and get the solution.

Complete step-by-step answer:
As we know that terminal digits in a number are only first and last digit.
So, let us fill the first and last place of the number with an even digit.
And we know that any digit is even if it is exactly divided by 2.
So, out of given digits there are only 3 even digits and that were {2, 4, 6}.
So, there are 3 possible digits that can be placed in the first place.
As we know that digits in a number are not repeated.
So, there will be 2 possible digits left for last place.
Now, we are left with 5 digits and 4 places. As we know that these four places can be filled with any number even or odd. But the digits cannot be repeated.
So, possible digits for second place will be 5.
Possible digits left for third place will be 4.
Possible digits left for fourth place will be 3.
And possible digits left for fifth place will be 2.
So, total number of numbers that can be formed with the digits {1, 2, 3, 4, 5, 6, 7} with no digits repeated and terminal digits as even will be 3$\times$5$\times$4$\times$3$\times$2$\times$2 = 720.
So, total possible numbers will be 720.
Hence, the correct option will be D.

Note: Whenever we come up with this type of problem then first, we will fill first and last place of the number with even digits and then other places. After placing a digit at any place, we subtract the number of digits left by 1. Because it is given that digits in a number cannot be repeated. And this will be the easiest and efficient way to find the solution of the problem.

The $7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 2$ would be the correct answer if we didn't have the final digit being even constraint. But as we do, it makes sense to think of the number in two separate parts: the last digit, and then the first five.

As you say, there are three choices for the terminal digit. But then, for the first five digits, we don't have a choice of seven digits any more - we have the choice of six [$\{1,3,5,7 \}$ or one of the unused two even digits]. So the number of options for the first five is $6\cdot 5\cdot 4\cdot 3\cdot 2 = 720$.

The final answer is therefore $3 \cdot 720 = 2160$ possible numbers.

EDIT: Just for fun, I confirmed the answer in Python, so the question can also be answered programmatically, though the above method is much preferable.

count = 0 for i in range[100000, 999999 + 1]: temp = list[int[x] for x in str[i]] if len[temp] == len[set[temp]] and temp[-1] % 2 == 0: if "0" not in str[i] and "8" not in str[i] and "9" not in str[i]: count += 1 print["count =", count] Output: count = 2160

here the question is how many 3 digit Even Numbers can be made using the digits 1 2 3 4 6 7 if no digit is repeated so we have total available for 1 2 3 4 6 and 7 so these are total 123456 station we have total 60 in which to 4 and 6 to 4 and 6 8 even digit now we need to form a three digit number exactly we need to form a three digit even number so we know that any one number is a number in which the last digit is even so we have we have three choices to fill this unit place

Índice

• Terminal digits mean the first digit and the last digit. Here, it is given that the terminal digits are even.So, The first position can be filled in 3 ways i.e, by 2,4 and 6Since the repetition is not allowed and the terminal digits need to be even, and we have used one even digit at the first position, So now we are left with '2‘ even digits, So, The last place can be filled in ’2' ways.Now, we have left with '5; digits and '4' places. As we know that these four places can be filled with any number even or odd. But the digits cannot be repeated.So, the possible digits for second place will be 5Possible digits left for third place will be 4Possible digits left for fourth place will be 3And possible digits left for the fifth place will be 2So, the total number of numbers that can be formed with the digits {1,2,3, 4,5,6,7} with no digits repeated and terminal digits as even=3×5×4×3×2×2=720So, the total possible numbers will be 720.
• The number of six-digit numbers that can be formed from the digits 1, 2, 3, 4, 5, 6 and 7 so that digits do not repeat, and the terminal digits are even isA. 144B. 72C. 288D. 720
• How many 6 digit even numbers can be formed from digits 1 2 3 4 5 7 and that the digit should not repeat and second last digit is even?
• How many 5 digit even numbers can be formed with 1 2 3 4 5 if the digits do not repeat?
• How many 6 digit even numbers can be made from the digits 2146 3 and 5?
• How many 6 digit even numbers can be formed from digits 1 to 7 so that the digits should not repeat and the second last digit is even?

now we already filled one position so we have total six digit so we have left with now 5 digit and the tenth place can be filled with 5 possibility since we have left 55 possibility because because the reputation is not allowed now we have already feel to position so we have left with four possibilities hence the hundred place can be filled with total 4 possibility Hands by multiplication total total number of 3 digit even numbers in which day in which repetition is not allowed on 4 into 5 into 3 if we simplify this way will get 60 hence the answer each 60

Solution

Terminal digits mean the first digit and the last digit. Here, it is given that the terminal digits are even.So, The first position can be filled in 3 ways i.e, by 2,4 and 6Since the repetition is not allowed and the terminal digits need to be even, and we have used one even digit at the first position, So now we are left with '2‘ even digits, So, The last place can be filled in ’2' ways.Now, we have left with '5; digits and '4' places. As we know that these four places can be filled with any number even or odd. But the digits cannot be repeated.So, the possible digits for second place will be 5Possible digits left for third place will be 4Possible digits left for fourth place will be 3And possible digits left for the fifth place will be 2So, the total number of numbers that can be formed with the digits {1,2,3, 4,5,6,7} with no digits repeated and terminal digits as even=3×5×4×3×2×2=720So, the total possible numbers will be 720.

The number of six-digit numbers that can be formed from the digits 1, 2, 3, 4, 5, 6 and 7 so that digits do not repeat, and the terminal digits are even isA. 144B. 72C. 288D. 720

Answer

Verified

Hint:Let us first fill the terminal digits of the number with even digits. After that fill all other places in a number. then find all the possible combinations and get the solution.

Complete step-by-step answer:
As we know that terminal digits in a number are only first and last digit.
So, let us fill the first and last place of the number with an even digit.
And we know that any digit is even if it is exactly divided by 2.
So, out of given digits there are only 3 even digits and that were {2, 4, 6}.
So, there are 3 possible digits that can be placed in the first place.
As we know that digits in a number are not repeated.
So, there will be 2 possible digits left for last place.
Now, we are left with 5 digits and 4 places. As we know that these four places can be filled with any number even or odd. But the digits cannot be repeated.
So, possible digits for second place will be 5.
Possible digits left for third place will be 4.
Possible digits left for fourth place will be 3.
And possible digits left for fifth place will be 2.
So, total number of numbers that can be formed with the digits {1, 2, 3, 4, 5, 6, 7} with no digits repeated and terminal digits as even will be 3$\times$5$\times$4$\times$3$\times$2$\times$2 = 720.
So, total possible numbers will be 720.
Hence, the correct option will be D.

Note: Whenever we come up with this type of problem then first, we will fill first and last place of the number with even digits and then other places. After placing a digit at any place, we subtract the number of digits left by 1. Because it is given that digits in a number cannot be repeated. And this will be the easiest and efficient way to find the solution of the problem.

Answer

Verified

Hint:Let us first fill the terminal digits of the number with even digits. After that fill all other places in a number. then find all the possible combinations and get the solution.Complete step-by-step answer:As we know that terminal digits in a number are only first and last digit.So, let us fill the first and last place of the number with an even digit.And we know that any digit is even if it is exactly divided by 2.So, out of given digits there are only 3 even digits and that were {2, 4, 6}.So, there are 3 possible digits that can be placed in the first place.As we know that digits in a number are not repeated. So, there will be 2 possible digits left for last place.Now, we are left with 5 digits and 4 places. As we know that these four places can be filled with any number even or odd. But the digits cannot be repeated. So, possible digits for second place will be 5.Possible digits left for third place will be 4.Possible digits left for fourth place will be 3.And possible digits left for fifth place will be 2.So, total number of numbers that can be formed with the digits {1, 2, 3, 4, 5, 6, 7} with no digits repeated and terminal digits as even will be 3$\times$5$\times$4$\times$3$\times$2$\times$2 = 720.So, total possible numbers will be 720.Hence, the correct option will be D. Note: Whenever we come up with this type of problem then first, we will fill first and last place of the number with even digits and then other places. After placing a digit at any place, we subtract the number of digits left by 1. Because it is given that digits in a number cannot be repeated. And this will be the easiest and efficient way to find the solution of the problem.

The $7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 2$ would be the correct answer if we didn't have the final digit being even constraint. But as we do, it makes sense to think of the number in two separate parts: the last digit, and then the first five.

As you say, there are three choices for the terminal digit. But then, for the first five digits, we don't have a choice of seven digits any more - we have the choice of six [$\{1,3,5,7 \}$ or one of the unused two even digits]. So the number of options for the first five is $6\cdot 5\cdot 4\cdot 3\cdot 2 = 720$.

The final answer is therefore $3 \cdot 720 = 2160$ possible numbers.

EDIT: Just for fun, I confirmed the answer in Python, so the question can also be answered programmatically, though the above method is much preferable.

count = 0 for i in range[100000, 999999 + 1]: temp = list[int[x] for x in str[i]] if len[temp] == len[set[temp]] and temp[-1] % 2 == 0: if "0" not in str[i] and "8" not in str[i] and "9" not in str[i]: count += 1 print["count =", count] Output: count = 2160

Q:

If it is possible to make a meaningful word with the first, the seventh, the ninth and the tenth letters of the word RECREATIONAL, using each letter only once, which of the following will be the third letter of the word? If more than one such word can be formed, give ‘X’ as the answer. If no such word can be formed, give ‘Z’ as the answer.

Answer & Explanation Answer: D] R

Explanation:

The first, the seventh, the ninth and the tenth letters of the word RECREATIONAL are R, T, O and N respectively. Meaningful word from these letters is only TORN. The third letter of the word is ‘R’.

View Answer Report Error Discuss

TCS Company Numerical Ability Permutation and Combination

• ------ given 6th digit even number , so last digit 2 or 4 or 6-> 3 ways " 5th digit should be even...so there will be 2 ways[rep. not allowed]

  so,therefore we get 5*4*3*2*2*3=720 ways

• 10 years agoHelpfull: Yes[61] No[17]
• there are six digits.. a b c d e f[say] f=possible digits are 3 -[2,4,6] e=possible 2 digits - [2,4,6] and then rest... d=possible 5 digits c=possible 4 digits b=possible 3 digits a=possible 2 digits the digits goes on decreasing since digits are not supposed 2 b repeated...

  hence ans=>3*2*5*4*3*2=720...:]

• 10 years agoHelpfull: Yes[33] No[4]
• _ _ _ _ _ _ last 2 digits should be even numbers. so 3p2 ways. remaining 4 digits should be filled with 5 numbers,becz no numbr is repeated. so 5P4 ways.

  ans=3P2*5P4=720

• 7 years agoHelpfull: Yes[16] No[1]
• 6th digit is filled by any one from 2,4,6 5th digit is placed by any 6 no, 4th digit is placed by any 5 no, 3rd digit is placed by any 4 no, 2nd digit is placed by any 3 no, 1st digit is placed by any 2 no, 2*3*4*5*6*3=2160
• 7 years agoHelpfull: Yes[10] No[10]
• second last digit is even. so, _ _ _ _ 2 _ _ _ _ _ 4 _ _ _ _ _ 6 _ 3 cases are possible. Now,1st vacant position can be filled up in 6 ways as 1 digit is fixed and we have to select it from rest 6 digits. Similarly 2nd position can be filled up in 5 ways 3rd position can be filled up in 4 ways 4th position can be filled up in 3 ways 6th position can be filled up in 2 ways So, total no of ways per case=6*5*4*3*2=720 ways.

  So, for 3 cases total no of ways= 3*720=2160

• 7 years agoHelpfull: Yes[7] No[0]
• let no. be a b c d e f then possible case for e=3 for f=2 possible case for a =5[total dig=7,1 used] for b=4 for c=3 for d=2 so 5*4*3*2*3*2=720
• 10 years agoHelpfull: Yes[6] No[3]
• ans is=3*6P5=2160
• 7 years agoHelpfull: Yes[6] No[7]
• there are six digits.. a b c d e f[say] f=possible digits are 3 -[2,4,6] e=possible 2 digits - [2,4,6] and then rest... d=possible 5 digits c=possible 4 digits b=possible 3 digits a=possible 2 digits the digits goes on decreasing since digits are not supposed 2 b repeated...

  hence ans=>3*2*5*4*3*2=720...:]

• 8 years agoHelpfull: Yes[3] No[0]
• Since it has been given to form 6 digit even number the last digit has to be one of the digits in either 2or 4 or 6 so its probability is 3P1 The condition given for the second last digit is" it should be the even number" since we have already placed one of the even number to the last digit from the remaining 2 digits we should fill the second last digit so 2P1 for the remaining places we are going to fill it with 4 digits among 5 because of the reason no repetition is allowed so 5P4 SOLUTION: 3*2*5*4*3*2 = 720

  Since all the condition has to be satisfied to get the 6 digit number.

• 7 years agoHelpfull: Yes[3] No[0]
• We have to form six digit numbers; so I will represent it as _ _ _ _ _ _ >Here the last two digits should be even >Out of the available 7 digits , 3 are even So the first 4 positions have 5*4*3*2 = 120 possibilities Now the last 2 positions can be filled in 3*2 = 6 ways

  Therefore the 6 digit no can be formed in 120*6=720 ways

• 8 years agoHelpfull: Yes[1] No[0]
• d is the ans
  5! x 3! =720
• 7 years agoHelpfull: Yes[1] No[0]
• Total kilometers travelled by 4 tyre = 40000 × 4 = 1,60,000. This has to be share by 5 tyres. So each tyre capacity = 1,60,000 / 5 = 32, 000. You have a doubt, after we travel 32,000 km, we are left with 4 worn tyes and one new tyre. But If the tyres are rotated properly after each 8000 km, all the tyres are equally used.
• 7 years agoHelpfull: Yes[1] No[0]
• Here they are talking about last two digit , that too should be in even . so consider even number from given question 2,4,6 possible ways are - - - - 24,- - - - 42,- - - - 46,- - - - 64,- - - -26,- - - - 62.

  So 6! ways=720 is answer

• 7 years agoHelpfull: Yes[1] No[2]
• 720
  5*4*3*2*2*3
• 6 years agoHelpfull: Yes[1] No[0]
• 720
  2x3x4x5x3x2 ways
• 4 years agoHelpfull: Yes[1] No[0]
• 3 * 6p5 =3*720=2160
• 7 years agoHelpfull: Yes[0] No[1]
• 5*4*3*2*2*3=720
  so answer is d]720
• 7 years agoHelpfull: Yes[0] No[1]
• 5*4*3*2*2*3=720
• 7 years agoHelpfull: Yes[0] No[2]
• we have to form 6 digits even number , _ _ _ _ _ _ so the last digit must be even. then only it is divisible by any even number here the condition is the last before digit must be also even there are three even numbers so the possibility of filling last digit is 3 and the possibility of filling before last digit is 2 remaining number including four odd number and 1 even number , therefore totally 5 the combination is 5*4*3*2*2*3=720
• 7 years agoHelpfull: Yes[0] No[0]
• 5!*2*3=720...
• 7 years agoHelpfull: Yes[0] No[2]
• ans.D] if we have to form even nmbers, unit's digit must be 2,4,6 i.e., 3 ways. Also 5th digit must be even so it can be filled in 2 ways . Now remaining 5 digits can be filled in 5! ways.

  so total 5!*3*2=720 ways.

• 7 years agoHelpfull: Yes[0] No[0]
• Ans is 720.
  
• 6 years agoHelpfull: Yes[0] No[0]
• answer is 720
• 6 years agoHelpfull: Yes[0] No[1]
• there are 6 digits. - - - - - - last digit should be even.[2,4,6] _ _ _ _ _ _3 ways. second last is even..there are 3 even numbers.we have to use one even number for last digit and remaining two even numbers for second last digit. _ _ _ _ _2 ways _3ways..

  5* 4*3*2*3=720 was.

• 6 years agoHelpfull: Yes[0] No[0]
• There are 7 digits and we have to form 6digit even nos. For a no to even , the last digit should be either[2,4,6] a/q the 2nd last digit should also be even [2,4,6] A/q no digits should repeat. therefore, last digit 3p1 second last digit 2p1 ___ ___ ___ ___ ___ ____[last both places are filled] first place can be filled by 5 digits remaining ,5p1 2nd=4p1 3rd =3p1 4th=2p1 on solving,

  5*4*3*2*2*3=720

• 4 years agoHelpfull: Yes[0] No[0]
• 6 digit even number so that the last digit will be 2 or 4 or 6 hence remaining 1357[246]
  totally 5! and the 2, 4, 6 has 3! so 5!*3!=1*2*3*4*5*1*2*3=720
• 4 years agoHelpfull: Yes[0] No[0]
• ------
  5*4*3*2*2*3=720
• 4 years agoHelpfull: Yes[0] No[0]
• To form 6-digit even number, the last digit should be an even number so 3 ways [2, 4, or 6] to fill the last digit and second last digit also should be even for which it will take 2 ways to fill.
  The last two digits are filled in 6 ways[ 2 x 3 = 6 ways]. The rest of the 4 digits can be filled in 5P4 ways i.e. 120 ways. Hence altogether to fill 6-digit even number = 120 * 6 = 720 ways.
• 4 years agoHelpfull: Yes[0] No[0]
• given 6 digit even taken as given number last digits filled with 2,4,6 in 3 ways next condition 5 th digit filled with 2 ways without repitition.

  so we get answer like 5*4*3*2*2*3 = 720

• 4 years agoHelpfull: Yes[0] No[0]
• 6P5*1*3=2160
• 3 years agoHelpfull: Yes[0] No[0]
• there are six digits.. a b c d e f[say] f=possible digits are 3 -[2,4,6] e=possible 2 digits - [2,4,6] and then rest... d=possible 5 digits c=possible 4 digits b=possible 3 digits a=possible 2 digits the digits goes on decreasing since digits are not supposed 2 b repeated...

  hence ans=>3*2*5*4*3*2=720.gowti

• 3 years agoHelpfull: Yes[0] No[0]
• 720
  5! x 4! x 3! x 2! x2! x3
• 19 Days agoHelpfull: Yes[0] No[0]

How many 6 digit even numbers can be formed from digits 1 2 3 4 5 7 and that the digit should not repeat and second last digit is even?

So, total number of numbers that can be formed with the digits {1, 2, 3, 4, 5, 6, 7} with no digits repeated and terminal digits as even will be 3×5×4×3×2×2 = 720. So, total possible numbers will be 720. Hence, the correct option will be D.

How many 5 digit even numbers can be formed with 1 2 3 4 5 if the digits do not repeat?

∴ the total number of arrangements = 120 + 96 + 96 = 312.

How many 6 digit even numbers can be made from the digits 2146 3 and 5?

This is Expert Verified Answer Therefore, There are 720 ways in which the number of 6 digit can be form by the number 214635.

How many 6 digit even numbers can be formed from digits 1 to 7 so that the digits should not repeat and the second last digit is even?

Therefore, 720, six digit even numbers can be formed.

How many 6 digit even numbers can be formed from digits 1 to 7 so that the digits should not repeat and the second last digit is even?

Therefore, 720, six digit even numbers can be formed.

How many 6 digit even numbers can be formed?

So, total number of numbers that can be formed with the digits {1, 2, 3, 4, 5, 6, 7} with no digits repeated and terminal digits as even will be 3×5×4×3×2×2 = 720. So, total possible numbers will be 720.

How many 6 digit numbers can be formed from 1 to 7 which are divisible by 5?

1 Answer. ∴ Required number of numbers divisible by 5 = 5 × 4 × 3 × 2 × 1 = 120.

How many six digit numbers can be formed using the digits 1 to 6 without repetition such that the number is divisible by the digit at its units place?

Hence, the correct answer is 720.

How many 6 digit even numbers can be formed from the digits 1 2 3 4 5 6 and 7 so that digits do not repeat and the second last digit is even?

How many 6 digit even numbers can be made from the digits 2146 3 and 5?

How many 3

How many six digit numbers can be formed using the digits 1 to 6 without repetition such that the number is divisible by the digit at its units place?