Quantitative Aptitudeमात्रात्मक योग्यता · क्रमचय और संचय
Permutation and Combinationक्रमचय और संचयquestions and answersप्रश्न और उत्तर
Permutation and combination is counting without listing. Almost every question comes down to two decisions: does order matter, and are any of the objects identical? Answer those two honestly and the formula picks itself.
Unlimitedअसीमितfresh questionsनए प्रश्न
Answersउत्तरalways one tap awayहमेशा एक टैप दूर
2subtopicsउप-टॉपिक
IBPS PO ~1
TCS NQT ~2
Practice questions are in English for now.अभ्यास के प्रश्न अभी अंग्रेज़ी में हैं।
How permutation and combination worksक्रमचय और संचय को समझिए
A permutation is an arrangement, so order matters: a president and a secretary chosen from 10 people can be filled in 10 × 9 = 90 ways. A combination is a selection, so order does not matter: a two-person team from the same 10 can be chosen in 90 ÷ 2 = 45 ways, because each team was counted twice.
Identical objects make some arrangements look the same. The letters of KERALA can be written in 6! = 720 orders, but swapping the two A’s never produces a new word, so the true count is 720 ÷ 2! = 360. Every letter that repeats gets its own factorial in the denominator.
Conditions are handled by shaping the count, not by patching it later. Letters that must stay together become one block; a person who must be included takes a seat first; and anything phrased as “at least one” is usually fastest as the total minus the “none” case. Around a round table one person can be fixed in place, which is why the count is (n − 1)! rather than n!.
Each trick says when to use it, why it works and when it fails. Formula tricks are checked by code over their whole range.हर ट्रिक बताती है कि उसे कब इस्तेमाल करें, वह क्यों काम करती है और कब नहीं। सूत्र वाली ट्रिक्स को कोड से उनकी पूरी रेंज पर जाँचा गया है।
Formulaसूत्रRarely taughtकम सिखाई जाने वाली✓ Checked on 5 cases✓ 5 मामलों पर जाँची गई
Two people never side by side at a round table: (n − 3) × (n − 2)!
Use it when:कब इस्तेमाल करें: n people sit around a round table and two particular people must not sit next to each other.
Write down (n − 2)!.
Multiply by (n − 3). That is the answer; no subtraction needed.
Exampleउदाहरण
8 friends at a round table; Asha and Ravi refuse to sit together.
→ (8 − 3) × 6! = 5 × 720 = 3,600.
Why it works, and when it failsयह क्यों काम करती है, और कब नहीं
Why:क्यों: All seatings: (n − 1)!. Seatings with the two together: treat them as one unit, (n − 2)! orders around the table, × 2 for their order = 2(n − 2)!. Subtracting, (n − 1)! − 2(n − 2)! = (n − 2)! × (n − 1 − 2) = (n − 3)(n − 2)!.
Limits:सीमाएँ: For n ≥ 3 people in a circle where only relative positions matter (a table, not a garland). In a row the count is n! − 2(n − 1)! = (n − 2)(n − 1)! instead.
Formulaसूत्र✓ Checked on 11 cases✓ 11 मामलों पर जाँची गई
Letters that must stay together: block outside × order inside
Use it when:कब इस्तेमाल करें: “In how many arrangements of the word … do the vowels (or some named letters) always come together?”
Tie the letters that must stay together into one block and count the units: the other letters + 1 block.
Arrange the units: (units)! ÷ (factorials of letters repeated outside the block).
Arrange the letters inside the block: (block size)! ÷ (factorials of letters repeated inside it). Multiply the two counts.
For “not all together”, subtract this from the total arrangements.
Exampleउदाहरण
Arrangements of COFFEE with all the vowels together.
→ Vowels O, E, E form a block; units C, F, F, block = 4 → 4!/2! = 12. Inside: 3!/2! = 3. Total 12 × 3 = 36.
Why it works, and when it failsयह क्यों काम करती है, और कब नहीं
Why:क्यों: Every arrangement with the vowels together is obtained exactly once by choosing an order for the units and then an order for the letters inside the block, so the two counts multiply (the multiplication principle). Repeated letters are divided out in whichever part they sit.
Limits:सीमाएँ: Only for one block of letters that must be adjacent. Two separate blocks each need their own inner count, and “no two vowels together” needs the gap method instead.
Formulaसूत्र✓ Checked on 195 cases✓ 195 मामलों पर जाँची गई
“At least one” = all − none
Use it when:कब इस्तेमाल करें: A committee, team or draw must contain “at least one” of some kind (a woman, a red ball, a senior member).
Count all selections with no restriction: C(n, k).
Count the selections with none of the required kind: C(others, k).
Subtract. For “at least two”, also subtract the selections with exactly one.
Exampleउदाहरण
A team of 3 from 5 men and 4 women with at least one woman.
→ C(9, 3) − C(5, 3) = 84 − 10 = 74.
Why it works, and when it failsयह क्यों काम करती है, और कब नहीं
Why:क्यों: Every selection either has no woman or has at least one, and never both, so the two counts add up to the total. Choosing “one woman, then anyone” instead counts a team with two women twice, once for each woman picked first.
Limits:सीमाएँ: Works whenever “none” is easy to count. If the condition is “at most” or “exactly”, count the allowed cases directly.
Solved examplesहल किए गए उदाहरण
Answers and full solutions are shown. Hide them to test yourself first.उत्तर और पूरे हल दिखाए गए हैं। पहले खुद को परखने के लिए इन्हें छिपाइए।
Example 1उदाहरण 1 · Arranging letters of a word and people in a circle · Easyआसान
Diya has letter tiles that spell PAPAYA. How many different arrangements of all 6 letters are possible?
10
60
120
720
Answer:उत्तर:60
PAPAYA has 6 letters: P twice, A three times.
Arrangements = 6! ÷ (2! × 3!) = 720 ÷ 12.
= 60.
Example 2उदाहरण 2 · Selections and committees with conditions · Easyआसान
In how many ways can 4 volunteers be chosen out of 15 for a cleanliness drive team?
Example 3उदाहरण 3 · Arranging letters of a word and people in a circle · Mediumमध्यम
Lakshmi rearranges the letters of PATNA. In how many of the arrangements do the vowels not all come together?
24
36
60
72
Answer:उत्तर:36
Total = 5! ÷ (2!) = 60.
Together: 4 units → 24; inside the block → 1; so 24.
Not all together = 60 − 24 = 36.
Example 4उदाहरण 4 · Selections and committees with conditions · Mediumमध्यम
From 5 men and 4 women, a panel of 3 is to be formed for a college fest. In how many ways can this be done if it must have exactly 1 woman?
14
40
80
84
Answer:उत्तर:40
Women: C(4, 1) = 4.
Men: C(5, 2) = 10.
Committees = 4 × 10 = 40.
Example 5उदाहरण 5 · Arranging letters of a word and people in a circle · Hardकठिन
7 cousins sit around a round table for a Baisakhi dinner in Ranchi. In how many different ways can they be seated? (Only their positions relative to one another matter.)
120
360
720
5,040
Answer:उत्तर:720
Fix one person. The other 6 can be arranged in 6! ways.
6! = 720.
Example 6उदाहरण 6 · Selections and committees with conditions · Hardकठिन
A committee of 3 is to be chosen from 12 people for a school in Patna. Tenzin and Rohan are among them and refuse to serve together. In how many ways can it be chosen?
10
100
120
210
Answer:उत्तर:210
All committees: C(12, 3) = 220.
Both Tenzin and Rohan in: C(10, 1) = 10.
Allowed = 220 − 10 = 210.
Common mistakesआम गलतियाँ
Using nPr for a committee or team: members of a team have no order, so divide by r!.
Forgetting to arrange the letters inside a block of “letters that must be together”.
Counting “at least one woman” as (choose one woman) × (choose the rest from everyone), which counts committees with two or more women several times.
Frequently asked questionsअक्सर पूछे जाने वाले प्रश्न
How do I know whether to use permutation or combination?
Ask whether swapping two chosen items gives a different outcome. If it does (ranks, posts, seats in a row, digits of a number), use permutation. If it does not (a team, a committee, a handful of fruit), use combination.
Why is the circular arrangement (n − 1)! and not n!?
Rotating everyone one seat to the left gives the same seating, since only positions relative to one another matter. Fixing one person removes those n rotations, leaving (n − 1)! distinct seatings. A garland can also be flipped over, so it is (n − 1)!/2.
What is the fastest way to handle “at least one” questions?
Count everything, then subtract the arrangements or selections that have none. For a committee of 4 from 6 men and 5 women with at least one woman: C(11, 4) − C(6, 4) = 330 − 15 = 315.