# Difficult Probability Solved QuestionAptitude Discussion

 Q. An eight-digit telephone number consists of exactly two zeroes. One of the digits is repeated thrice. Remaining three digits are all distinct. If the first three digits (from left to right) are 987, then find the probability of having only one 9, one 8 and one 7 in the telephone number.
 ✖ A. 1/18 ✖ B. 1/20 ✔ C. 1/10 ✖ D. 5/47

Solution:
Option(C) is correct

Case A: There is only one 9, one 8 and one 7 in the number.

Hence, there has to be one digit from {1,2,3,4,5,6} repeated thrice.

Total number of ways in which such a number can exist:

$={^6C_1}\times \dfrac{5!}{3!\times 2!}$

$=60$

Case B: One of the three digit {9,8,7} is repeated thrice.

Hence there will be one digit from {1,2,3,4,5,6}.

Total number of ways in which such a number can exist:

$={^3C_1} \times {^6C_1}\times \dfrac{5!}{3!\times 2!}$

$=540$

Total possible telephone numbers:

$=60+540=600$

Probability = 60/600 = 1/10

## (10) Comment(s)

Akshay
()

Elaborate Method -

Given that first three positions of the eight positions is filled with 9,8 and 7 we have remaining 5 positions to be filled with numbers with two zeros and three numbers which meet the given conditions. Let's consider four cases -

Case 1 - Suppose the number that is repeated thrice is 9. We already have one 9 which leaves us with two 0's and two 9's in the five vacant positions. Remaining one position can take values from 1 to 6. So

$6C1*(5!/(2!*3!))=6*30=180$

Case 2 - Similar to case 1 but here considering the number repeating thrice is 8. So totally 180 combinations.

Case 3 - Similar to case 1 but here considering the number repeating thrice is 7. So totally 180 combinations.

Case 4 - Considering the first three numbers 9,8 and 7 as distinct numbers, the three vacant spaces has to be filled with three same numbers which can range from 1 to 6 and the other two vacant position has to take two 0's. So

$6C1*1C1*1C1*(5!/(2!*3!))=6*10=60$

So totally we have $(180*3)+60=600$ combinations. Now coming to the question it asks for the case which has one 9, one 8 and one 7 which is the case 4. So probability is

$60/600=1/10$

Divya
()

In case B the total no. Of ways should be 180+60= 240

Rashmi
()

have u assumed the positions of zeros fixed????????//

()

I guess since there are two zeros present, their position is NOT assumed here to be fixed but as a repeated number. That is why we see a factor of $2!$ in the denominator in both the cases.

Isn't it?

Bibek Sahu
()

The answer must be $\dfrac{3c1*6c1*(5!/(2!*2!))}{3c1*6c1*(5!/(2!*2!))}+60$

Samyak Jain
()

its 60/(3c1*6c1*5!/(2!*2!)+60)

Anil Kumar
()

Danalc
()

agree with crujzo.

Hardik
()

In case B,

3.6.5!/(2!2!)

looks right. How does it have 3! instead of 2?

Crujzo
()

I believe in the second case it should be,

$2c1 * 6c1 * (5! / (2!*2!))$

As from the rest of the left 5 positions, there are 2 & 2 repeated items