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01 Nov 2019, 05:47
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
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55% (02:35) correct
45%
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wrong
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We are required to form different words with the help of letters of the word INTEGER. Let m1 be the number of words in which I and N are never together and m2 be the number of words which begin with I and end with R, then what is the value if m1/m2?
A. 18
B. 24
C. 30
D. 36
E. 40
Are You Up For the Challenge: 700 Level Questions
A. 18
B. 24
C. 30
D. 36
E. 40
Are You Up For the Challenge: 700 Level Questions
01 Nov 2019, 06:21
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Bunuel
Formula: Not together = Total ways - no of ways when together
Number of ways when I&N together = arranging the letters {IN, T, E, G, E, R} in a straight line = 6!*2!/2! = 6! = 720
Total ways of arranging INTEGER = 7!/2! = 2520
Not together, m1 = 2520 - 720 = 1800
Number of ways which begin with I and end with R is same as arranging the letters {N, T, E, G, E} in the given spaces I, _, _, _, _, _, R = 5!/2! = 60 = m2
m1/m2 = 1800/60 = 30
IMO Option C
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General Discussion
01 Nov 2019, 07:55
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Bunuel
INTEGER can be arranged in 7!/2! ways = 2520
M1 = m1 be the number of words in which I and N are never together = INTEGER ; XTEGER = 6!/2!*2c1 ; 720
so M1 = 2520-720 ; 1800
M2 = m2 be the number of words which begin with I and end with R ; I NTEGE R ; 5!/2! ; 60 ways
ratio M1/M2 = 1800/60 ; 30
IMO C
25 Mar 2020, 08:04
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Bunuel
INT-EGE-R has 7 letters, 2 identical (E), so 6 distinct
p(INTEGER)=7!/2!
P([IN]TEGER)=6!/2!*2C1 (because we can begin with [IN] or [NI])
P(without [IN])=7!/2!-6!=(7!-2*6!)/2!=(6!(7-2))/2!)=6!(5)/2!
P([I]NTEGE[R])=5!/2!
m1/m2=6!(5)/2!/5!/2!=6!(5)/2!*2!/5!=6*5=30
Ans (C)
