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25 Nov 2019, 01:20
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A
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Difficulty:
75%
(hard)
Question Stats:
57% (02:32) correct
43%
(02:34)
wrong
based on 143
sessions
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All the possible five-digit numbers are formed using only the digits 4, 5 and 6. How many of them will have at least two fives?
(A) 131
(C) 195
(B) 219
(D) 230
(E) 250
(A) 131
(C) 195
(B) 219
(D) 230
(E) 250
25 Nov 2019, 08:17
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CaptainLevi
Total cases \((3)^4\) - ( Ways to get no five + ways to get one five ) = Ways to get at least two five.
No five's =2*2*2*2*2 = 32
Number of ways to get ONLY one 5:
One five and rest all 4 ( 5, 4, 4, 4, 4) = \(\frac{5!}{4!}= 5\) ways
One five and rest all 6 ( 5, 6, 6, 6, 6) = \(\frac{5!}{4!}= 5\) ways
One five ,three 6 and one 4 ( 5, 6, 6, 6, 4) =\(\frac{5!}{3!}=20\) ways
One five ,three 4 and one 6( 5, 4, 4, 4, 6)=\(\frac{5!}{3!}\)=20 ways
One five ,two 6 and two 4 ( 5 6 6 4 4) =\(\frac{5!}{2!*2!}\)=30 ways
So total ways=\((3)^4\)=243
No fives in = 32 ways
One five in = 5+5+20+20+30 =80 ways
So 243 - (32+80)= 131 Ways to get at least two five's.
Ans- A
