### Discussion :: Permutation and Combination - General Questions (Q.No.10)

Gayathri said: (Apr 30, 2011) | |

Why we use permutation concept instead of combination. |

Manasa said: (Jun 14, 2011) | |

@Gayathri In this case we are not selecting any letters. Just we are arranging the letters according to given condition. |

Kesav said: (Jul 21, 2011) | |

Can any one solve this problem for me? 1! + 2! + .... + 50!=? a) 3.1035*10^64 b) 2.1021*10^65 c) 3.1035*10^63 d) 3.1035*10^62 |

Riddhi said: (Jul 24, 2011) | |

@kesav:plz give me ans. Is it b? |

Vishnu said: (Aug 14, 2011) | |

It is asked in TCS apti test. |

Gautam said: (Dec 28, 2011) | |

There are 6 letters in the given word, out of which there are 3 vowels and 3 consonants. Number of ways of arranging the vowels = nPr= n!/(n-r)= 3/(3-3) =3!/0! ,=3*2*1= 6 {we define 0! = 1} Also, the 3 consonants can be arranged at the remaining 3 positions. Number of ways of arranging the vowels = nPr= n!/(n-r)= 3/(3-3) =3!/0! ,=3*2*1= 6 {we define 0! = 1}. Total number of ways = (6 x 6) = 36. |

Nikhil said: (Jan 30, 2013) | |

Vowel can even be put at the 7th place, not considered above, which will increase the number of possible arrangements. Please clarify? |

Athu said: (Apr 11, 2013) | |

What is odd positions? |

Drushya said: (Apr 20, 2013) | |

@Athu. Odd position means its like odd no. For example : 1 2 3 4 5 6. In this case odd position is :1 3 5 & even position : 2 4 6. |

Jinal said: (Jul 28, 2013) | |

Vowels can also take position 7 (the last position) ? V C V C V C V. So the answer would be 3!*4C3 = 24. Is this correct? |

Vasudha said: (Jul 31, 2013) | |

Since there is nothing mentioned about repetitions so the no ways could also be 3*3*3(i.e., E can come in all three odd places and same goes for the other two vowels A and I as well) for the three odd positions of vowels. And similarly the other three letters can be arranged so the answer should be 3*3*3*3*3*3. |

Sanjay said: (Sep 1, 2013) | |

Hi guys. There is 3 odd, 3 even place and the word have 3 vowel and 3consonant. We arrenge 3 vowel in 3 odd place and 3 consonant in 3 rest place such that VCVCVC So total no of way = 3p3*3p3 = 36. |

Neha Pant said: (Aug 17, 2014) | |

But here it is not mentioned that repetition is allowed so according to me the correct way is, 3*3*3*3*3*3 = 729 which is not in option. 3*3*3 for 3 vowels and other for consonant . Please explain it? |

Nee said: (Oct 16, 2014) | |

Can't we do it in this way like first we find all possible ways of arranging all letters. i.e 6! and then subtract 4!(3!) from 6! to get the answer. |

Priya said: (Feb 7, 2015) | |

Any example of cards combination. Tell me. |

Prakash S said: (Jul 13, 2015) | |

I really don't understand why we are doing that. How we are taken 3p3 = 6? |

Rohini said: (Jul 24, 2015) | |

Why write this type 3p3? |

Subhadeep said: (Aug 23, 2015) | |

We can consider the seventh position also. So I feel the answer is wrong. |

Raji said: (Aug 24, 2015) | |

I think we arrange 3 consonants + 1 vowel = 4 (generally 3 vowels but total vowels take as 1 set). So 4! = 12. The three vowels re arrange is 3!=6. So answer is 72. Is there any wrong, please tell me. |

Sourav Sarkar said: (Sep 1, 2015) | |

Well, there are four odd positions according to me. 1234567. 1, 3, 5, 7 are the odd positions. 3 vowels have to be selected to fit in any of the three position out of four. This can be done in 4C3 ways. The rest remains the same. Answer- 4C3x3!x3! |

Shivam said: (Sep 3, 2015) | |

There are 4 possible position for the vowel so why 4C3 is not included in the solution for selecting 3 position out of 4 for vowels? |

Balakrishna said: (Sep 4, 2015) | |

Here they didn't mention about repetition allowed or not if we do this problem with repetition allowed the answer may be not 36. |

Devdp said: (Nov 24, 2015) | |

When 1st try myself consider x vowel and y consonants it can be possibility of position (1) xyxyxy and (2) yxyxyx. As per permutation for x 3p3 = 3! = 6. Permutation for why 3y3 = 3! = 6. And as shown as above 2 different position. Total arrangement = 6*6*2 = 72 (Ans). But when I shows option I confused. There is no 72. |

Devdp said: (Nov 25, 2015) | |

In 3rd line permutation for why 3p3 = 3! = 6. This small mistake done bye me in writing but answer I got is 72. Reply anyone. |

Divyansh said: (Nov 27, 2015) | |

Brother only odd position so answer is 36. |

Dev said: (Dec 22, 2015) | |

It can be done by 6x5x4x3x2 divide by 2 and again divide by 10 = 36. |

Pinky Kumari said: (Aug 10, 2016) | |

The answer will be 36. It is correct. Because there is 3 odd place for vowel and 3 for a consonant. |

Aditya said: (Aug 12, 2016) | |

How can we solve the same question taking 'FATHER' as the given word? |

M223 said: (Aug 21, 2016) | |

How many ways can the word DETAIL be rearranged such that no two vowels are placed adjacently? Please tell me the answer. |

Aman said: (Sep 1, 2016) | |

Why making this question so complicated? See we have the word "DETAIL" which is having the total number of alphabets that is "6". In which "3" is vowels and "3" is consonants and we have to arrange the vowels at odd places right so what are the positions we have (1) 2 (3) 4 (5) 6 [number inside the braces indicates the odd places]. So we have 3 places and we need to arrange the 3 vowels we do it by combination that is 3c3=1 and the no of vowels is 3 so their permutation is 3! so that the similar permutation of consonants is having the same 3! because we have 3 consonants as well so finally the answer is => 3C3 * 3! * 3! => 1* 6 * 6. => 36. |

Aman said: (Sep 1, 2016) | |

@Aditya. I have the solution for your question you have the word "FATHER" that is nothing but 6 letters in which 2 vowels and 4 consonants. So we have to arrange the vowels at odd places as we know that out of 6 words 3 places are odd places, so the arrangement of 2 vowels at 3 places means 3C2 right. So we arranged that vowel at odd places now we have to take the permutation of that vowels as 2! and the permutation of the consonants that is nothing but 4! so finally we have the solution as :- => 3C2 * 2! * 4! => 3 * 2 * 24. => 144. |

Kat@93 said: (Sep 12, 2016) | |

@Devdp. I too agree with you. If I name the places from left-hand side i.e. 6-5-4-3-2-1 or from right-hand side 1-2-3-4-5-6. I will get 2 types of arrangement and since the question is regarding permutation both way of arrangement matters, so answer we get is 72. |

Gopika S S said: (Sep 28, 2016) | |

There are 6 positions out of which 1, 3 and 5 are for the three vowels. There are 3 ways for arranging in position 5 and remaining 2 ways for position 3 and 1 way for position 1. So for vowels, it is 3 * 2 * 1 = 6. For remaining three consonants which are placed in even position, there are 3 ways in position 6 and remaining 2 ways in position 4 and one way in position 2. So for consonants, it is 3 * 2 * 1 = 6. So a total number of ways = 6 * 6 = 36. |

Tommy said: (Oct 6, 2016) | |

It is 36. Assuming we represent the filings as D*T**L. So that d (*) the possible fillings for the vowels (E A I). Then "E" can go into the arrangement in 3ways I. E either "E" takes d first, second, or the third filling space ie shown by the (*) sign. So after E as occupying one space, then we are left with just spaces left which are to be filled by "A" and "I". In which "A" can go into it in 2ways. Remaining one space to be occupied by "I" in just 1way. So arrangement of vowel is 3P3 = 3! = 3 * 2 * 1 = 6. So we have 3 consonant (D T L) left we can permute (arrange) them in 3! ways = 3 * 2 * 1 = 6 ways. So total arrangement = 6 * 6 = 36 ways. |

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