Top 50 Most Repeated PERMUTATIONS AND COMBINATIONS PYQs | JEE MAINS
A curated collection of the most important questions from PERMUTATIONS AND COMBINATIONS, fully solved with step-by-step concepts to prepare for JEE MAINS.
A curated collection of the most important questions from PERMUTATIONS AND COMBINATIONS, fully solved with step-by-step concepts to prepare for JEE MAINS.
Treat the pair as one unit: $4!$ arrangements. Pair can switch: $\times2$. Total: $3! \times 2 = 12$...
Read Full Step-by-Step Solution →Fix one boy, arrange remaining 5 boys in 5! ways, 6 gaps for girls in 6! ways. Total = $5! \times 6! = 86400$....
Read Full Step-by-Step Solution →Arrange 5 boys in circle: $4!$. Girls in gaps: $5!$. Total: $4! \times 5! = 576$....
Read Full Step-by-Step Solution →Letters: E, H, M, O, R, T. Fix positions: Words starting with E, H → 5! each = 120. Then M: next letter E → 4! = 24, H → 24, O: now 'OT' → T then H → ...
Read Full Step-by-Step Solution →Use inclusion-exclusion: $3^5 - 3\cdot2^5 + 3\cdot1^5 = 243 - 96 + 3 = 150$...
Read Full Step-by-Step Solution →Use permutation formula $^5P_3 = 5!/(5-3)! = 5×4×3$....
Read Full Step-by-Step Solution →Simplify using $6! = 720$, $5! = 120$, $4! = 24$: $(720 - 600)/24 = 5$ → Wait: recalculate: $6! = 720$, $5\cdot5! = 600$, diff=120,
Use multinomial expansion; count combinations giving $x^4$....
Read Full Step-by-Step Solution →Treat the pair as one unit: $4!$ arrangements. Pair can switch: $\times2$. Total: $3! \times 2 = 12$...
Read Full Step-by-Step Solution →Fix one boy, arrange remaining 5 boys in 5! ways, 6 gaps for girls in 6! ways. Total = $5! \times 6! = 86400$....
Read Full Step-by-Step Solution →Arrange 5 boys in circle: $4!$. Girls in gaps: $5!$. Total: $4! \times 5! = 576$....
Read Full Step-by-Step Solution →Letters: E, H, M, O, R, T. Fix positions: Words starting with E, H → 5! each = 120. Then M: next letter E → 4! = 24, H → 24, O: now 'OT' → T then H → ...
Read Full Step-by-Step Solution →Use inclusion-exclusion: $3^5 - 3\cdot2^5 + 3\cdot1^5 = 243 - 96 + 3 = 150$...
Read Full Step-by-Step Solution →Use permutation formula $^5P_3 = 5!/(5-3)! = 5×4×3$....
Read Full Step-by-Step Solution →Simplify using $6! = 720$, $5! = 120$, $4! = 24$: $(720 - 600)/24 = 5$ → Wait: recalculate: $6! = 720$, $5\cdot5! = 600$, diff=120,
Use multinomial expansion; count combinations giving $x^4$....
Read Full Step-by-Step Solution →Treat the pair as one unit: $4!$ arrangements. Pair can switch: $\times2$. Total: $3! \times 2 = 12$...
Read Full Step-by-Step Solution →Fix one boy, arrange remaining 5 boys in 5! ways, 6 gaps for girls in 6! ways. Total = $5! \times 6! = 86400$....
Read Full Step-by-Step Solution →Arrange 5 boys in circle: $4!$. Girls in gaps: $5!$. Total: $4! \times 5! = 576$....
Read Full Step-by-Step Solution →Letters: E, H, M, O, R, T. Fix positions: Words starting with E, H → 5! each = 120. Then M: next letter E → 4! = 24, H → 24, O: now 'OT' → T then H → ...
Read Full Step-by-Step Solution →Use inclusion-exclusion: $3^5 - 3\cdot2^5 + 3\cdot1^5 = 243 - 96 + 3 = 150$...
Read Full Step-by-Step Solution →Use permutation formula $^5P_3 = 5!/(5-3)! = 5×4×3$....
Read Full Step-by-Step Solution →Simplify using $6! = 720$, $5! = 120$, $4! = 24$: $(720 - 600)/24 = 5$ → Wait: recalculate: $6! = 720$, $5\cdot5! = 600$, diff=120,
Use multinomial expansion; count combinations giving $x^4$....
Read Full Step-by-Step Solution →Treat the pair as one unit: $4!$ arrangements. Pair can switch: $\times2$. Total: $3! \times 2 = 12$...
Read Full Step-by-Step Solution →Fix one boy, arrange remaining 5 boys in 5! ways, 6 gaps for girls in 6! ways. Total = $5! \times 6! = 86400$....
Read Full Step-by-Step Solution →Arrange 5 boys in circle: $4!$. Girls in gaps: $5!$. Total: $4! \times 5! = 576$....
Read Full Step-by-Step Solution →Letters: E, H, M, O, R, T. Fix positions: Words starting with E, H → 5! each = 120. Then M: next letter E → 4! = 24, H → 24, O: now 'OT' → T then H → ...
Read Full Step-by-Step Solution →Use inclusion-exclusion: $3^5 - 3\cdot2^5 + 3\cdot1^5 = 243 - 96 + 3 = 150$...
Read Full Step-by-Step Solution →Use permutation formula $^5P_3 = 5!/(5-3)! = 5×4×3$....
Read Full Step-by-Step Solution →Simplify using $6! = 720$, $5! = 120$, $4! = 24$: $(720 - 600)/24 = 5$ → Wait: recalculate: $6! = 720$, $5\cdot5! = 600$, diff=120,
Use multinomial expansion; count combinations giving $x^4$....
Read Full Step-by-Step Solution →Treat the pair as one unit: $4!$ arrangements. Pair can switch: $\times2$. Total: $3! \times 2 = 12$...
Read Full Step-by-Step Solution →Fix one boy, arrange remaining 5 boys in 5! ways, 6 gaps for girls in 6! ways. Total = $5! \times 6! = 86400$....
Read Full Step-by-Step Solution →Arrange 5 boys in circle: $4!$. Girls in gaps: $5!$. Total: $4! \times 5! = 576$....
Read Full Step-by-Step Solution →Letters: E, H, M, O, R, T. Fix positions: Words starting with E, H → 5! each = 120. Then M: next letter E → 4! = 24, H → 24, O: now 'OT' → T then H → ...
Read Full Step-by-Step Solution →Use inclusion-exclusion: $3^5 - 3\cdot2^5 + 3\cdot1^5 = 243 - 96 + 3 = 150$...
Read Full Step-by-Step Solution →Use permutation formula $^5P_3 = 5!/(5-3)! = 5×4×3$....
Read Full Step-by-Step Solution →Simplify using $6! = 720$, $5! = 120$, $4! = 24$: $(720 - 600)/24 = 5$ → Wait: recalculate: $6! = 720$, $5\cdot5! = 600$, diff=120,
Use multinomial expansion; count combinations giving $x^4$....
Read Full Step-by-Step Solution →Treat the pair as one unit: $4!$ arrangements. Pair can switch: $\times2$. Total: $3! \times 2 = 12$...
Read Full Step-by-Step Solution →Fix one boy, arrange remaining 5 boys in 5! ways, 6 gaps for girls in 6! ways. Total = $5! \times 6! = 86400$....
Read Full Step-by-Step Solution →Arrange 5 boys in circle: $4!$. Girls in gaps: $5!$. Total: $4! \times 5! = 576$....
Read Full Step-by-Step Solution →Letters: E, H, M, O, R, T. Fix positions: Words starting with E, H → 5! each = 120. Then M: next letter E → 4! = 24, H → 24, O: now 'OT' → T then H → ...
Read Full Step-by-Step Solution →Use inclusion-exclusion: $3^5 - 3\cdot2^5 + 3\cdot1^5 = 243 - 96 + 3 = 150$...
Read Full Step-by-Step Solution →Use permutation formula $^5P_3 = 5!/(5-3)! = 5×4×3$....
Read Full Step-by-Step Solution →Simplify using $6! = 720$, $5! = 120$, $4! = 24$: $(720 - 600)/24 = 5$ → Wait: recalculate: $6! = 720$, $5\cdot5! = 600$, diff=120,
Use multinomial expansion; count combinations giving $x^4$....
Read Full Step-by-Step Solution →Treat the pair as one unit: $4!$ arrangements. Pair can switch: $\times2$. Total: $3! \times 2 = 12$...
Read Full Step-by-Step Solution →Fix one boy, arrange remaining 5 boys in 5! ways, 6 gaps for girls in 6! ways. Total = $5! \times 6! = 86400$....
Read Full Step-by-Step Solution →