TCS Company Numerical Ability Permutation and Combination

Find the number of ways in which one or more letter be selected from the letters AAAABBCCCDEF.
A. 479
B. 480
C. 958
D. 960

Read Solution (Total 4)

As A occurs 4 times, we can choose A in 4+1 ways

So, the number of ways of selection of no or more letters from the word:
AAAABBCCCDEF will be N=(4+1)(2+1)(3+1)(1+1)(1+1)(1+1)
Clearly, this contains one combination of choosing no letter

So, the number of ways of selection of one or more letters will be N−1
ans=480-1=479 (A)
9 years agoHelpfull: Yes(40) No(2)

((4+1)(2+1)(3+1)(1+1)(1+1)(1+1))-1 = 479
9 years agoHelpfull: Yes(5) No(2)
2^12-1/4!3!2!
why it is wrong?
9 years agoHelpfull: Yes(1) No(3)
@TEJASVITA nice interpretation but u are considering 2^12 for every number two possibilities , we are not taking factorial eg 2^4=8 only to divide it should be 4!=24.hope it vl clear ur doubt little bit.
9 years agoHelpfull: Yes(1) No(1)

TCS Other Question

WHAT IS UNIT DIGIT OF 2^(3^28439)? In a sequence of integers, A(n) = A(n-1) - A(n-2), where A(n) is the n th term in the sequence, n is an integer and n >=3, A(1)=1,A(2)=1 . Calculate S(1000), where S(1000) is the sum of the first 1000 terms