Let x = 787,422. In how many ways can the digits in x be rearranged

Question

Let x = 787,422. In how many ways can the digits in x be rearranged so that the result is greater than x?

A. 20
B. 30
C. 60
D. 120
E. 180

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8.toastcrunch · Answer

Let fix 8 at first and then we left with 5 digits and after than 5!/2!2! = 30,
Answer is  .

fdfd97 · Answer

The only possible arrangement to get a number greater than 787422 is any number more with 8 at the left most spot.
Also, there are 2 repetitions of 7 and 2.
So, no. of possibilities= (1 * 5!)/(2!*2!) = 30 - Option B

mysterymanrog · Answer

First: 
The number is 787,422. This is a 6 digit number. 
For the rearrangement to be higher than the intial X, one possible way is to have the digit 8 be first. In that case, the rest of the digits can be in any order. 
So, 8(ab,cde) is one possibility. The total number of ways is 1*5!, and then dividing that by 2!*2!, since we have a repeat 7 and a repeat 2. That gives us a total of 30 possibilities 
Another option would be to have the number 7 at the start. However, if you do this, it will be impossible for the rearrangement to be higher. If 7 is first, then 8 must be second - but the remaining digits (7,4,2,2) can only be arranged in a way such that they are equal or less than the original number. So we can completely exclude this case.

The final answer is B.

Archit3110 · Answer

x = 787422
for value of digits re arranged such that is greater than x
877422
(77422) can be arranged in 5! / (2!2!) ways or say 30
option B

bumpbot · Answer

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Let x = 787,422. In how many ways can the digits in x be rearranged

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GMAT Problem Solving (PS) Questions

Let x = 787,422. In how many ways can the digits in x be rearranged

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