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CAT 2020 Slot 2 — QA Question 3

Numbers (P&C)Easy

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

Answer & solution

Answer: 315

Solution

Easy

The digits 77 and 33 are fixed; choose the other two digits, count all arrangements, then use symmetry — in exactly half of all such numbers 77 precedes 33. Split into cases by whether 00 is among the chosen digits, since a 44-digit number cannot start with 00.

The four digits are 77, 33, and two more chosen from {0,1,2,4,5,6,8,9}\{0,1,2,4,5,6,8,9\}. "77 before 33" means 77 appears in an earlier position than 33; by symmetry this happens in half of all arrangements.

1

Case 1: 00 is not chosen. Pick 22 more digits from the 77 nonzero non-{3,7}\{3,7\} digits, then arrange all four freely.

(72)=21(choices of the two digits)4!=24(arrangements, no leading-zero issue)⇒ 21×24=504 numbers\begin{aligned} &\binom{7}{2} = 21 \quad\text{(choices of the two digits)}\\ &4! = 24 \quad\text{(arrangements, no leading-zero issue)}\\ &\Rightarrow\ 21\times 24 = 504\ \text{numbers} \end{aligned}

By symmetry 77 precedes 33 in half of them.

5042=252\frac{504}{2} = 252
2

Case 2: 00 is chosen. Pick 11 more digit from the remaining 77. Among 4!4! arrangements, those with leading 00 are invalid: the first place has 33 valid digits, the rest 3!3!.

(71)=7(choice of the extra digit)3×3!=18(valid arrangements, no leading 0)⇒ 7×18=126 numbers\begin{aligned} &\binom{7}{1} = 7 \quad\text{(choice of the extra digit)}\\ &3\times 3! = 18 \quad\text{(valid arrangements, no leading }0\text{)}\\ &\Rightarrow\ 7\times 18 = 126\ \text{numbers} \end{aligned}

Again 77 precedes 33 in half.

1262=63\frac{126}{2} = 63
3

Total. Add the two cases (from steps 1 and 2).

252+63=315252 + 63 = 315
315315
CAT 2020 Slot 2 QA Q3: How many 4-digit numbers, each greater than 1000 and each having all four digits distinct, are there with 7 co — Solution | TheCATExam