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## Friday, 30 November 2012

## Wednesday, 28 November 2012

### Problem Solving : Permutation and combination Problem 3: If the letter of the word SEQUESTERED are arranged in all possible ways and these words are written out as in a dictionary form, then what is the 50th rank of the word SEQUESTERED is ?

**If the letter of the word SEQUESTERED are arranged in all possible ways and these words are written out as in a dictionary form, then what is the 50th rank of the word SEQUESTERED is ?**

Solution:

So the word

Hence the number of words begin with

and the number of words begin with

**SEQUESTERED**can be arranged in dictionary form of letters as**D E E E E Q R S S T U**Hence the number of words begin with

**DEEEEQR = 4! / 2! =12**and the number of words begin with

**DEEEEQS = 4! =24**

the number of words begin with**DEEEEQT = 4! / 2! =12**

**12+24+12 =48**

so next word

hence next words are 49 and 50 will be

DEEEEQURSST is 49th word

and

**U**becomes last wordhence next words are 49 and 50 will be

DEEEEQURSST is 49th word

and

**DEEEEQURSTS is the 50th word**

and the answer -->and the answer -->

**50th rank of the word SEQUESTERED is "****DEEEEQURSTS****"**### Problem Solving : Permutation and combination Problem 2: If the letter of the word VERMA are arranged in all possible ways and these words are written out as in a dictionary form, then what is the rank of the word VERMA is ?

If the letter of the word

so the word VERMA letters of

In alphabetical order

and the number of words begin with

and the number of words begin with

and the number of words begin with

after the word begins with

So the number of words begin with

**VERMA**are arranged in all possible ways and these words are written out as in a dictionary form, then what is the rank of the word**VERMA**is ?

**Solution:**so the word VERMA letters of

**V, E, R, M, A**In alphabetical order

**A, E, M, R, V****So the number of words begin with**

**A = 4! (leave A and consider remaining letters E, M, R, V)**and the number of words begin with

**E****= 4! (leave A and consider remaining letters A, M, R, V)**and the number of words begin with

**M****= 4! (leave A and consider remaining letters E, A, R, V)**and the number of words begin with

**R****= 4! (leave A and consider remaining letters E, M, A, V)**after the word begins with

**V**so here after consider two or more words combinationsSo the number of words begin with

**VA****= 3! (leave v and A and consider remaining 3 letters E,M, R)****VE**becomes next character**after VA**set so below are the dictionary arrangement goes**1) VEAMR****2) VEARM**

3) VEMAR

4) VEMRA3) VEMAR

4) VEMRA

**5) VERAM**

finally 6) VERMA

Hence the Rank = 4! + 4! + 4! + 4! + 3! + 6 = 96 + 6+ 6 =108th Rank

Rank of the word VERMA is 108.finally 6) VERMA

Hence the Rank = 4! + 4! + 4! + 4! + 3! + 6 = 96 + 6+ 6 =108th Rank

Rank of the word VERMA is 108.

## Monday, 26 November 2012

### Problem Solving : Permutation and combination Problem 1 : In how many ways can the letter of the word PROPORTION be arranged by taking 4 letters at a time?

In how many ways can the letter of the word

below I've wrote letters without repeated words and then how many times particular word repeated

from the question here we gonna select 4 words hence we have the following chance to select those 4 words

Hence

**PROPORTION**be arranged by taking 4 letters at a time?**Solution:**

**P R O P O R T I O N**below I've wrote letters without repeated words and then how many times particular word repeated

**P R O T I N****P R O****O**

**Here (P, P) 2 P, (R, R) 2 R, and (O,O,O) 3 O**from the question here we gonna select 4 words hence we have the following chance to select those 4 words

**Chance 1:**3 O and remaining 1 are different (That means (O,O,O)--> 3 word and (P,R,T,I,N) ---> 1 word from the 5)**Chance 2:**two of same words and other two of the same words (here (O,O,O), (T,T) and (P,P) out of 3 pairs we gonna select 2 pair of words )**Chance 3:**Two of same words and other two are different words (here (O,O,O), (T,T) and (P,P) out of 3 pairs we gonna select 1 pair of word and other 2 word from (R, O, I, N, T) words consider P,P has been selected pair )**Chance 4:**All 4 are different ( here P, R, O, T, I , N out of 6 we gonna select 4 words)Hence

**Chance 1: 3C3 * 5C1 * (4! /3!) = 20 [3! for repeated words O,O,O]**

Chance 2: 3C2 * (4! / (2!*2!)) = 18 [2! * 2! for repeated words of T,T and P,P]

Chance 3: 3C1*5C2 * (4! / 2!) = 360 [2! for repeated word T,T or P,P]

Chance 4: 6C4 * 4! = 360 [as usual]

finally add all the chances = 20 + 18 + 360 + 360 = 758 ways is the answerChance 2: 3C2 * (4! / (2!*2!)) = 18 [2! * 2! for repeated words of T,T and P,P]

Chance 3: 3C1*5C2 * (4! / 2!) = 360 [2! for repeated word T,T or P,P]

Chance 4: 6C4 * 4! = 360 [as usual]

finally add all the chances = 20 + 18 + 360 + 360 = 758 ways is the answer

## Friday, 16 November 2012

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