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CAT Previous Year Paper | CAT Quants Questions | Question 26

Combinatorics has confused and frightened the students since years! If you find yourself struggling with this topic, remember that you are not alone! This topic haunts the best of us! The best way to come out of this fright is to face it! And one of the best ways to do so is to put yourself through questions which we have seen in the CAT Question Paper. Solve CAT previous year paper and get comfortable with this tricky topic. Don't keep it for later, start practicing now! Here is a question on combinatorics from CAT 2020 question paper! See if you can sovle it without looking at the solution.

Question 26 : How many 4-digit numbers, each greater than 1000 and each having all four digits distinct, are there with 7 coming before 3?


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Explanatory Answer

_ _ _ _ on the 4 digit 7 has to come before 3
So, 4C2 possibilities for 7 coming before 3 into the spaces
7 ___ 3 ____, 2 numbers are already there and 2 remaining spots we have to fill in
From 8 digits available (Excluding 7 and 3)
So, 8C2 ways of choosing
After choosing two from 8, that two can be placed in any way
For example : 1 and 2 can be arranged as 12 and 21
So, 4C2 × 8C2 × 2
Now we need to subtract the possibility where 0 comes in the first position
0753, So these are in the form 0__ __ __
In this 3 places 7 before 3 can be placed in 3C2 ways and remaining 1 digit
Can be chosen from 7 (Excluding 0,3,7) digits.
3C2 × 7 = 21 numbers
Should subtract 21 from 4C2 × 8C2 × 2
= (6 × 28 × 2) – 21
= 315


The question is "How many 4-digit numbers, each greater than 1000 and each having all four digits distinct, are there with 7 coming before 3?"

Hence, the answer is, " 315"

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