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Find the product of (1-1/6)(1-1/7)(1-1/8)................(1-1/n+5) for n>=4 (a)1/n+5 (b)2/n+4 (c)5/n+5 (d)3/(n+4)(n+5) (e)3/n+3 Ans :c (1-1/6)(1-1/7)(1-1/8)……(1-1/(n+5)) =(6/6-1/6)(7/7-1/7)(8/8-1/8)…… ((n+5)/(n+5)-1/(n+5)) =(5/6)(6/7)(7/8)……((n+4)/(n+5)) Cancel 6,7,8,……,(n+4): =5/(n+5)

Complete the series: 4,3,5,9,12,17,________ (A)29 (B)25 (C)26 (D)27 Ans:26 4+5= 9 3+9=12 5+12=17 So, 9+17=26

In a party of 80 people each person handshakes with the other. Find the total number of handshakes? A)3160  B)3280 C)3260  D)2296 Answer:A)3160  To calculate the number of handshakes, we can use the formula n(n-1)/2, where n is the number of people. In this case, n = 80. So, the number of handshakes would be 80(80-1)/2 = 3,160 handshakes.

seven people are seated in a row. probability that two particular persons A &B are seated at the end is? (a) 1/14      (b)1/21            (c)1/81           (d)2/15 ANS: To find the probability that two particular persons, A and B, are seated at the end of a row when seven people are seated in total, we can consider the total number of ways the seven people can be arranged and then calculate the number of favorable outcomes where A and B are seated at the end. Total number of ways to arrange 7 people in a row = 7! Now, when A and B are seated at the end, they can be arranged among themselves in 2! ways. Additionally, the remaining 5 people can be arranged in 5! ways. Therefore, the number of favorable outcomes where A and B are seated at the end = 2! * 5! The probability that A and B are seated at the end = (Number of favorable outcomes) / (Total number of outcomes) = (2! * 5!) / 7! Calculating the values: 2!...

In a given code SISTER is coded as 535301. UNCLE as 84670 and BOY as 129. How is RUSTIC written in the code? (A) 633185              (B) 185336               (C) 363815              (D) 581363 ANS: From given code we obtain the following  result S=5        I=3          S=5          T=3           E=0       R=1 U=8       N=4          C=6         L=7         E=0 R=1     U=8            S=5          T=3         I=3        C=6

Find the number ways of arranging 5 boys and 6 girls around a circular table, such that (1)no two boys sit together ANS: First to arrange 6 girls in round table (6-1)!=120ways 6 place is available in between girls to sit boys and 5 boys available it can be done by 6P5 ways Total no of ways=6!*6P5=120*120=14400

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