Latest Elitmus Aptitude Question SOLUTION: Q. How many 5 letter words can be formed from ANDHRA PRADESH such that it should start with letter A and end with letter R.

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(#M40030986) ELITMUS Numerical Ability question PERMUTATION Keep an EYE

Q. How many 5 letter words can be formed from ANDHRA PRADESH such that it should start with letter A and end with letter R.

Asked In Elitmus pawan kumar (10 years ago)

Unsolved Read Solution (15)

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Read Solution (15) :

ravi  (10 years ago)

A (2A 2D 2H NPESR)R
1)there is only 1 way of selecting 1 out of 3 identical A.similarly way for R
2)now we have to select 3 words out of these remaining (2A 2D 2H NPESR)
2.a) two similar and 1 different word: ---->two similar can be selected from 2A 2D 2H --->3C1
Now we have to select 1 word out of remaining either
(2A 2D NPESR)or (2D 2H NPESR) or (2A 2H NPESR) --->7C1
So we have 3C1 X 7C1,but there can be arrangement among these two similar and 1 different among themselves.------> 3C1 X 7C1 X (3!/2!) =63
2.b) all three are different---> 8C3 X 3! = 336
So,336 +63=399 is the right answer

Is this solution Helpfull? Yes (27) | No (6)  

K.ARun  (10 years ago)

AAA(NDHPDESH)RR,
SO 3c1*8c3*2c1
= 3*56*2
336

Is this solution Helpfull? Yes (9) | No (15)  

Kunal Wadhwa  (8 years ago)

We have to fix first and last position as A and R respectively
A _ _ _ R

Now we have left with following letters :
A -> 2
H ->2
D -> 2
N -> 1
P -> 1
R -> 1
E -> 1
S -> 1

So two cases can occur :
Case 1: When all three letters are different
A _________ _________ _________ R
(8 ways) * (7 ways) * (6 ways) = 336 ways

Case 2 : When two letters are alike
Suppose two alike letters are D
A D D __ R or A D __ D R or A __ D D R
(7 ways) + (7 ways) + (7 ways) = 21 ways

And we similar cases with 2 H as well as with 2 A. So total 63 (21*3) such cases.

So overall we have 336+63 = 399 cases possible

Is this solution Helpfull? Yes (9) | No   

Ajay Sharma  (10 years ago)

Ans:- 381

Sol:- After placing a 'A' at 1st position and a 'B' at the last we are now left with A-2 ,D-2 ,H-2 ,N-1 ,R-1 ,P-1 ,E-1 ,S-1 so now there are 2 cases

Case 1: 2 alike and 1 different
Selection for above case is (3C1 * 5C1)
And these selected letters can be arranged in (3! / 2!) ways
So total possibilities in this case is = 3C1 * 5C1 * (3! / 2!)
= 45

Case 2: all three different
Selection for above case is 8C3
And these selected letters can be arranged in 3! ways
So total possibilities in this case is = 8C3 * 3!
= 336

So total ways = 336+45
= 381

Is this solution Helpfull? Yes (7) | No (7)  

Nandha  (10 years ago)

No struggle in solving this pbm in tedious way. its answer is 990. first letter begins with 'A' and end with 'R', The question all so didn't state any constraint such as without repeatation or meaningful, so there are remaining 11 lettersto fill in the three places(A_ _ _ R). so the answer is 11P3=990.

Is this solution Helpfull? Yes (5) | No (1)  

ashes  (10 years ago)

:- After placing a 'A' at 1st position and a 'B' at the last we are now left with A-2 ,D-2 ,H-2 ,N-1 ,R-1 ,P-1 ,E-1 ,S-1 so now there are 2 cases

Case 1: 2 alike and 1 different
AA_ IMPLIES 7*3!/2! HERE 7 FOR SEVEN DIFF LETTERS AND 3!/2! FOR ARRANGEMENTS
SIMILARLY FOR DD_ AND HH_. HENCE TOTAL PERMUTATIONS FOR CASE 1 ARE (7*3!/2!)*3=63.

CASE 2: ALL DIFFERENT (_ _ _)NOW TOTAL PERMUTATIONS ARE 8P3=336.
ADDING CASE1 AND CASE 2 WE GET 399 WAYS.

Is this solution Helpfull? Yes (4) | No (3)  

Simran  (10 years ago)

The first letter is A and the last one is R.
therefore one has to find three more letters from the remaining 11 letters.
Of the 11 letters, there are 2 Ds and 2Hs and 2As and one each of the remaining 5 letters.
The second, third and fourth positions can either have three different letters or have same.
Case 1: When the three letters are different. One has to choose three different letters from the available different choices. This can be done in 8 * 7*6 = 336 ways.
Case 2: When the three letters are same. There are 3 options - the three can be either As or Hs or Ds. Therefore 3 ways.
Total number of possibilities = 336 + 3 = 339

Is this solution Helpfull? Yes (2) | No (2)  

Anushi saxena  (10 years ago)

i think it is 990
11c3*3c1*2c1=990

Is this solution Helpfull? Yes (1) | No (5)  

mahima srivastava  (10 years ago)

the word will be a_ _ _r
so a can be selected in 3c1 ways
r can be selected in 2c1 ways
and rest three letters in 11c3 ways
now we have to arrange three letters. we have 2a ,2d,2h and rest of the letters one time each
so, there are two possibilities either two of them will be same or all of them will be different

therefore no. of ways=(3c1*11c3*2c1)*((3!/2!)+(3!))=8910

Is this solution Helpfull? Yes (1) | No (6)  

pawan kumar  (10 years ago)

ans. plz m waiting......

Is this solution Helpfull? Yes | No (2)  

Ajay Sharma  (10 years ago)

Ans:-1389 ways

Is this solution Helpfull? Yes | No (8)  

P DEEPAK KUMAR  (10 years ago)

i also have the answer 336...... suggest if wrong

Is this solution Helpfull? Yes | No (3)  

vishal rathore  (10 years ago)

this is the question that made me feel that i have confusion in permutation and combination .. :P but still lets solve this A and R are fixed hence remaining three places is filled in 8p3 ways , ithink so 336 is answer as we have to fill remaining three places with 11 letters but ,AA ,HH ,DD are counted as 1 :)

Is this solution Helpfull? Yes | No (3)  

rohitashwa  (10 years ago)

11c3/3*2!==165

Is this solution Helpfull? Yes | No   

Digesh Singh Bisht  (1 year ago)

(11!)/2!*2!*2!

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