Latest TCS Aptitude Question SOLUTION: find the total number of combinations of 5 letters a,b,a,b,b taking some or all at a time?

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(#M40021889) TCS Numerical Ability question Keep an EYE

find the total number of combinations of 5 letters a,b,a,b,b taking some or all at a time?

Asked In TCS reddy siri (10 years ago)

Unsolved Read Solution (15)

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Read Solution (15) :

monika suhag  (10 years ago)

total no of ways of taking some or all out of p+q things such that p are of one type and q are of another type is (p+1)(q+1)-1
so,ans is a comes 2 times i.e. p=2
b comes 3 times i.e q=3
ans =(2+1)(3+1)-1
=11

Is this solution Helpfull? Yes (54) | No (5)  

VIKAS SINGH  (10 years ago)

first 1 digit no. can be a or b(2 ways 2 select), for 2 digit no.fallowing comb.
r possible..aa,ab,ba,bb(4ways)
for 3 digit no's lets 3b,2b nd 1a,1b nd 2a.
so total ways 1+3+3=7ways.
for 4 digit no's 3b nd 1a,2b nd 2a,
so total ways 4+6=10ways(coz using 3b nd 1a..4!/3!=4 nd 4!/2!*2!=6).
for 5 digit no.s 3b nd 2a.
so total ways=5!/3!*2!=10
so total combination of no.s(1digit+2digit+3digit+4digit+5digit)..
2+4+7+10+10=33
so ans will be 33.

Is this solution Helpfull? Yes (20) | No (21)  

sanjib paul  (10 years ago)

5!/(2!*3!)=10.
so! 10 is d ri8 ans.

Is this solution Helpfull? Yes (8) | No (5)  

Uma shankar Pradhan  (8 years ago)

1 letter can be chosen in 2 ways. a or b
2 letters can be chosen in 3 way. aa, ab, bb
3 letters can be chosen in 3 ways. bbb, aab, bba
4 letters can be chosen in 2 ways. aabb, bbba
5 letters can be chosen in 1 way.
So total ways are 11

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

pavan  (10 years ago)

5!/(2!*3!)=20

Is this solution Helpfull? Yes (5) | No (15)  

sums  (10 years ago)

if we consider a , b as a factors then the question reflects as factors of a given number
so we get a^2*b^3
no of factors we can get from formula (2+1)*(3+1)=12
in factors 1 is also a factor .. here there is no such case (i.e not selecting any)
so 12-1 = 11 will be the answer

Is this solution Helpfull? Yes (4) | No (1)  

Prianka  (8 years ago)

The idea is to find not the combinations but the freq of letters.
ababb
2a's 3b's 1 letter stuffs a,b=2,2 letter stuffs aa,bb,ab=3,3letter stuffs aab,bbb,abb=3,4 letter stuffs=abbb,aabb=2,5 letter stuffs=1 adding these up we get 2+3+3+2+1= 11ways

Is this solution Helpfull? Yes (2) | No   

Rajanikanta Pradhan  (8 years ago)

11ways is correct answer

Is this solution Helpfull? Yes (2) | No   

Girdhari Lal Choudhary  (8 years ago)

1 letter can be chosen in 2 ways. a or b
2 letters can be chosen in 3 way. aa, ab, bb
3 letters can be chosen in 3 ways. bbb, aab, bba
4 letters can be chosen in 2 ways. aabb, bbba
5 letters can be chosen in 1 way.
So total ways are 11

Is this solution Helpfull? Yes (2) | No   

hemanth kumar  (10 years ago)

(5c1+5c2+5c3+5c4+5c5)(3!*2!)=31

Is this solution Helpfull? Yes (1) | No (8)  

reddy siri  (10 years ago)

correct ans given is 11..but dnt know how to solve

Is this solution Helpfull? Yes (1) | No (10)  

akshat sinha  (8 years ago)

Ans is 11
The best and detailed answer was given here..
http://linkshrink.net/7dq5SJ

Is this solution Helpfull? Yes (1) | No (7)  

Rudra  (10 years ago)

I think exact ans is 15.
5p1+5p5/2!*3!
so by solving it ans is 15.

Is this solution Helpfull? Yes | No (7)  

Shubham gupta  (7 years ago)

i think it should be 10

Is this solution Helpfull? Yes | No   

Aman  (7 years ago)

1 letter can be chosen in 2 ways. a or b
2 letters can be chosen in 3 way. aa, ab, bb
3 letters can be chosen in 3 ways. bbb, aab, bba
4 letters can be chosen in 2 ways. aabb, bbba
5 letters can be chosen in 1 way.
So total ways are 11

Is this solution Helpfull? Yes | No (1)  

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