(1) In how many ways can you select at least one chocolate, one pastry and one biscuit from 2 chocolate,3 pastries and 4 biscuits?
(a) 350 (b)315 (c)250 (d) 275
(2)1!+2!+3!+…………………………….+68!.what is the digit at tenth place?
(a)0 (b)1 (c)5 (d) 7

(3)In a sequence of numbers,the sum of the first n terms is equal to 11n^2+10n.What is the sum of the 2nd,4th and 6th in the sequence?
(a)250 (b)300 (c)261 (d) cant say

• ans of 1st is 315 ways
  bcz no.of ways of selecting 1 or more items frm n items is 2^n-1
  apply dis
  for chocolate 2^2-1=3
  for pastries 2^3-1=7
  for biscuits 2^4-1=17
  now 3*7*15=315
  
• 11 years agoHelpfull: Yes(21) No(0)
• sum of first n terms=11n^2+10n
  sum of first(n-1) terms=11(n-1)^2+10(n-1)
  calculate it =11n^2-12n+1
  nth term=sum of n terms-sum of n-1 terms
  nth term=22n-1
  hence find 2,4 6 term by putting n=2,4.....
  ans =261
• 11 years agoHelpfull: Yes(11) No(0)
• -->(3) 1st n terms sum = n(11n+10)
  then calculate for 1st term we get = 21
  ----"""------------2nd------------ = 43
  -------------------3.......=65
  ------4------=87
  ---5-----=199
  ----6-----=131
  then sum of 2nd + 4th + 6th =261
  Ans: 261
• 11 years agoHelpfull: Yes(5) No(0)
• 2nd one :-
  as the no. of zeros from 10! - 68! consist of two or more zeros at two last places thus now we have to consider only 1! + 2! + ...+ 9! =....13 where .... represents some value. Thus digit at tenth place is 1. Ans :1
• 11 years agoHelpfull: Yes(3) No(3)
• (3) s(n) = 11n^2+10n
  s(n-1) = 11n^2 -12n +1
  As we know that
  an=s(n) - s(n-1)=22n-1
  thus we get values as s2 s4 s6 as 43 87 131 thus suming we get 261
• 11 years agoHelpfull: Yes(2) No(0)
• i dont know
• 11 years agoHelpfull: Yes(1) No(10)
• average of 1st n numbers=11n+10.......(1)
  averafe of 2nd,3rd,4th number=sum(2nd+3rd+4th)/3........(2)
  [sum(2nd+3rd+4th)/3]+(n-3)=11n+10
  sum(2nd+3rd+4th)+3n-9=33n+30
  sum(2nd+3rd+4th)=30n+39
• 11 years agoHelpfull: Yes(0) No(3)
• nmKDO GOkg nlkg n
• 4 years agoHelpfull: Yes(0) No(0)