Skip to content
Miscellaneous Exercise · Q4

Q.How many 4 digit numbers can be formed from the digits 1, 1, 2, 2, 3, 3, 4 and 5?

Chandigarh CbseNCERTSubjective· 3mImportance★★★★★est
93% · 117/126 Questions
🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

The 8 available digits are 1,1,2,2,3,3,4,51,1,2,2,3,3,4,5 — digits 1,2,31,2,3 can each appear at most twice, 44 and 55 at most once — so split the count into cases by how many digits repeat in the 4-digit number.

Number of distinct arrangements of a multiset of kk items with repeated groups of sizes p1,p2,…p_1,p_2,\ldots is k!p1! p2!⋯\dfrac{k!}{p_1!\,p_2!\cdots}. Combine with nCr=n!r!(n−r)!^nC_r=\dfrac{n!}{r!(n-r)!} to first choose which digit-values are used.

  1. Case A — all four digits distinct. Choose any 4 distinct digit-values out of the 5 available values {1,2,3,4,5}\{1,2,3,4,5\}: 5C4=5^5C_4=5 ways. Arrange the 4 distinct digits: 4!=244!=24 ways. Sub-total =5×24=120=5\times24=120.
  2. Case B — exactly one digit repeated twice, other two digits distinct. The repeated digit must be one that has 2 copies available, i.e. from {1,2,3}\{1,2,3\}: 33 choices. Choose the other 2 (distinct, non-repeated) digit-values from the remaining 4 values: 4C2=6^4C_2=6 ways. Arrange the multiset {x,x,y,z}\{x,x,y,z\} (one pair + two singles) in 4!2!=12\dfrac{4!}{2!}=12 ways. Sub-total =3×6×12=216=3\times6\times12=216. …

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.