Muft Shiksha™ एक 100% Free Education Portal है 🇮🇳, जिसका उद्देश्य Class 9–12 के हर विद्यार्थी तक High-Quality Education को पूरी तरह मुफ्त पहुँचाना है। 🇮🇳 हम मानते हैं कि अच्छी शिक्षा किसी student की आर्थिक स्थिति पर निर्भर नहीं होनी चाहिए। 🇮🇳 हर विद्यार्थी को वही Quality Study Material, MCQs, Quizzes, Exam Preparation, Concept-Based Learning और Bilingual Support मिलना चाहिए, जो आमतौर पर महंगी Coaching या Premium Platforms में मिलता है। Muft Shiksha™ 🇮🇳 इसी सोच के साथ बनाया गया है • Muft Shiksha™ एक 100% Free Education Portal है 🇮🇳, जिसका उद्देश्य Class 9–12 के हर विद्यार्थी तक High-Quality Education को पूरी तरह मुफ्त पहुँचाना है। 🇮🇳 हम मानते हैं कि अच्छी शिक्षा किसी student की आर्थिक स्थिति पर निर्भर नहीं होनी चाहिए। 🇮🇳 हर विद्यार्थी को वही Quality Study Material, MCQs, Quizzes, Exam Preparation, Concept-Based Learning और Bilingual Support मिलना चाहिए, जो आमतौर पर महंगी Coaching या Premium Platforms में मिलता है। Muft Shiksha™ 🇮🇳 इसी सोच के साथ बनाया गया है • Muft Shiksha™ एक 100% Free Education Portal है 🇮🇳, जिसका उद्देश्य Class 9–12 के हर विद्यार्थी तक High-Quality Education को पूरी तरह मुफ्त पहुँचाना है। 🇮🇳 हम मानते हैं कि अच्छी शिक्षा किसी student की आर्थिक स्थिति पर निर्भर नहीं होनी चाहिए। 🇮🇳 हर विद्यार्थी को वही Quality Study Material, MCQs, Quizzes, Exam Preparation, Concept-Based Learning और Bilingual Support मिलना चाहिए, जो आमतौर पर महंगी Coaching या Premium Platforms में मिलता है। Muft Shiksha™ 🇮🇳 इसी सोच के साथ बनाया गया है
A. सबसे बड़े चुने हुए element को fix करना/Fixing the largest chosen element
Step 1
Concept
This is the hockey-stick identity and the largest selected element creates cases. In exams use the last-element method for such staircase sums.
Step 2
Why this answer is correct
The correct answer is A. सबसे बड़े चुने हुए element को fix करना / Fixing the largest chosen element. This is the hockey-stick identity and the largest selected element creates cases. In exams use the last-element method for such staircase sums.
Step 3
Exam Tip
यह hockey-stick identity है और सबसे बड़ा selected element cases बनाता है। परीक्षा में ऐसी staircase sums में last element method लगाएं।
B. इसे hockey-stick identity का shifted रूप मानना/Treat it as a shifted form of the hockey-stick identity
Step 1
Concept
The upper and lower indices increase together so it is a diagonal sum. In exams think of hockey-stick when a diagonal combination sum appears.
Step 2
Why this answer is correct
The correct answer is B. इसे hockey-stick identity का shifted रूप मानना / Treat it as a shifted form of the hockey-stick identity. The upper and lower indices increase together so it is a diagonal sum. In exams think of hockey-stick when a diagonal combination sum appears.
Step 3
Exam Tip
Upper और lower indices साथ बढ़ रहे हैं इसलिए यह diagonal sum है। परीक्षा में diagonal combination sum दिखे तो hockey-stick सोचें।
B. एक subset चुनकर उसमें unordered marked pair चुनना/Choosing a subset and selecting an unordered marked pair inside it
Step 1
Concept
Choose the marked pair first and choose the remaining elements freely. In exams think of pair marking when \({}^{r}C_2\) appears.
Step 2
Why this answer is correct
The correct answer is B. एक subset चुनकर उसमें unordered marked pair चुनना / Choosing a subset and selecting an unordered marked pair inside it. Choose the marked pair first and choose the remaining elements freely. In exams think of pair marking when \({}^{r}C_2\) appears.
Step 3
Exam Tip
पहले marked pair चुनें और बाकी elements freely choose करें। परीक्षा में \({}^{r}C_2\) दिखे तो pair marking सोचें।
Choose the (3) marked members first and freely choose the remaining (n-3) members. In exams count the inner combination first.
Step 2
Why this answer is correct
The correct answer is C. \(^{n}C_3 2^{n-3}\). Choose the (3) marked members first and freely choose the remaining (n-3) members. In exams count the inner combination first.
Step 3
Exam Tip
पहले (3) marked members चुनते हैं और बाकी (n-3) members freely चुने जाते हैं। परीक्षा में inner combination को पहले count करें।
If a pair is marked and then one member is marked, cases occur inside or outside the pair. In exams split products of marks into cases.
Step 2
Why this answer is correct
The correct answer is D. \(2{}^{r}C_2+3{}^{r}C_3\). If a pair is marked and then one member is marked, cases occur inside or outside the pair. In exams split products of marks into cases.
Step 3
Exam Tip
पहले pair mark हो और फिर एक member mark हो तो cases pair के अंदर या बाहर बनते हैं। परीक्षा में product of marks को cases में तोड़ें।
Choose the innermost (c) elements first and then add layers. In exams count nested choices from inside to outside.
Step 2
Why this answer is correct
The correct answer is A. \(^{n}C_c{}^{n-c}C_{b-c}{}^{n-b}C_{a-b}\). Choose the innermost (c) elements first and then add layers. In exams count nested choices from inside to outside.
Step 3
Exam Tip
सबसे अंदर के (c) elements पहले चुनें और फिर layers जोड़ें। परीक्षा में nested choices को अंदर से बाहर count करें।
The smaller selected set must lie inside the larger selected set. In exams first check valid index order in nested combinations.
Step 2
Why this answer is correct
The correct answer is A. \(s\leq r\leq n\). The smaller selected set must lie inside the larger selected set. In exams first check valid index order in nested combinations.
Step 3
Exam Tip
छोटा selected set बड़े selected set के अंदर होना चाहिए। परीक्षा में nested combination में valid index order पहले check करें।
After the second derivative and putting (x=-1), ((1-1)^{n-2}) appears. In exams check zero when the falling factor is below the degree.
Step 2
Why this answer is correct
The correct answer is C. (0). After the second derivative and putting (x=-1), ((1-1)^{n-2}) appears. In exams check zero when the falling factor is below the degree.
Step 3
Exam Tip
Second derivative के बाद (x=-1) रखने पर ((1-1)^{n-2}) आता है। परीक्षा में degree से कम falling factor हो तो zero check करें।
Using (r-3=r(r-1)(r-2)+3r(r-1)+r) gives this form. In exams use falling factorial decomposition for cubic sums.
Step 2
Why this answer is correct
The correct answer is A. (n-2(n+3)2^{n-3}). Using (r-3=r(r-1)(r-2)+3r(r-1)+r) gives this form. In exams use falling factorial decomposition for cubic sums.
Step 3
Exam Tip
(r-3=r(r-1)(r-2)+3r(r-1)+r) से यह form मिलता है। परीक्षा में cubic sums में falling factorial decomposition लगाएं।
Two groups have the same size (a), so interchanging them gives duplicates. In exams divide extra by factorial for equal-size unlabelled groups.
Step 2
Why this answer is correct
The correct answer is C. (2!). Two groups have the same size (a), so interchanging them gives duplicates. In exams divide extra by factorial for equal-size unlabelled groups.
Step 3
Exam Tip
दो groups का size (a) समान है इसलिए उनकी अदला-बदली duplicate देती है। परीक्षा में equal-size unlabelled groups पर extra factorial divide करें।
Both internal orders and equal-size group swaps are removed. In exams write both group sizes and equal-group factorials in the denominator.
Step 2
Why this answer is correct
The correct answer is A. (3!3!4!4!2!2!). Both internal orders and equal-size group swaps are removed. In exams write both group sizes and equal-group factorials in the denominator.
Step 3
Exam Tip
Internal orders और equal-size group swaps दोनों हटते हैं। परीक्षा में denominator में group sizes और equal-group factorials दोनों लिखें।
A. (^{r}C_1\left((r-1)^n-{}^{r-1}C_1(r-2)^n+\cdots\right))
Step 1
Concept
Choose the empty box first and then distribute onto the remaining (r-1) boxes. In exams use choose empty plus onto count for exactly non-empty boxes.
Step 2
Why this answer is correct
The correct answer is A. (^{r}C_1\left((r-1)^n-{}^{r-1}C_1(r-2)^n+\cdots\right)). Choose the empty box first and then distribute onto the remaining (r-1) boxes. In exams use choose empty plus onto count for exactly non-empty boxes.
Step 3
Exam Tip
पहले empty box चुनें और बाकी (r-1) boxes में onto distribution करें। परीक्षा में exactly non-empty boxes में choose empty plus onto count करें।
Choose the empty boxes first and then the remaining two boxes must both be non-empty. In exams apply onto to the remaining boxes in exactly-empty cases.
Step 2
Why this answer is correct
The correct answer is A. (^{4}C_2\(2^n-2\)). Choose the empty boxes first and then the remaining two boxes must both be non-empty. In exams apply onto to the remaining boxes in exactly-empty cases.
Step 3
Exam Tip
पहले empty boxes चुनें फिर बाकी दो boxes दोनों non-empty होने चाहिए। परीक्षा में exactly empty में remaining boxes पर onto लगाएं।
Empty-box cases are removed by inclusion-exclusion. In exams count non-empty labelled boxes like onto functions.
Step 2
Why this answer is correct
The correct answer is A. \(3^n-3\cdot2^n+3\). Empty-box cases are removed by inclusion-exclusion. In exams count non-empty labelled boxes like onto functions.
Step 3
Exam Tip
Inclusion-exclusion से empty box cases हटते हैं। परीक्षा में non-empty labelled boxes को onto functions की तरह गिनें।
After removing the minimum sum (10), (12) remains and is distributed among (4) variables. In exams subtract lower bounds and use stars and bars.
Step 2
Why this answer is correct
The correct answer is B. \(^{15}C_3\). After removing the minimum sum (10), (12) remains and is distributed among (4) variables. In exams subtract lower bounds and use stars and bars.
Step 3
Exam Tip
Minimum sum (10) हटाने पर (12) बचता है और (4) variables में distribute होता है। परीक्षा में lower bounds subtract करके stars and bars लगाएं।
The upper violation is \(x_i\geq8\) and inclusion-exclusion is applied. In exams use shift (8) for upper bound (7).
Step 2
Why this answer is correct
The correct answer is A. \(^{21}C_2-3{}^{13}C_2+3{}^{5}C_2\). The upper violation is \(x_i\geq8\) and inclusion-exclusion is applied. In exams use shift (8) for upper bound (7).
Step 3
Exam Tip
Upper violation \(x_i\geq8\) है और inclusion-exclusion applied होता है। परीक्षा में upper bound (7) के लिए shift (8) लें।
Choose the zero variable and split positive sum (16) among the remaining (3) variables. In exams convert exactly-zero cases into positive distribution.
Step 2
Why this answer is correct
The correct answer is A. \(^{4}C_1{}^{15}C_2\). Choose the zero variable and split positive sum (16) among the remaining (3) variables. In exams convert exactly-zero cases into positive distribution.
Step 3
Exam Tip
Zero variable चुनें और बाकी (3) variables में positive sum (16) बांटें। परीक्षा में exactly zero cases को positive distribution में बदलें।
First choose the (3) non-empty boxes and then split (18) balls into (3) positive parts. In exams use choose boxes plus positive stars and bars for exactly non-empty.
Step 2
Why this answer is correct
The correct answer is A. \(^{6}C_3{}^{17}C_2\). First choose the (3) non-empty boxes and then split (18) balls into (3) positive parts. In exams use choose boxes plus positive stars and bars for exactly non-empty.
Step 3
Exam Tip
पहले (3) non-empty boxes चुनें फिर (18) balls को (3) positive parts में बांटें। परीक्षा में exactly non-empty को choose boxes plus positive stars-bars करें।
B. जब पहला object और उसका target object एक दूसरे की जगह बदल लें/When the first object and its target object swap places
Step 1
Concept
If two objects swap with each other, the remaining (n-2) objects are deranged. In exams separate swap and non-swap cases in derangement recurrence.
Step 2
Why this answer is correct
The correct answer is B. जब पहला object और उसका target object एक दूसरे की जगह बदल लें / When the first object and its target object swap places. If two objects swap with each other, the remaining (n-2) objects are deranged. In exams separate swap and non-swap cases in derangement recurrence.
Step 3
Exam Tip
यदि दो objects आपस में swap करते हैं तो बाकी (n-2) objects derange होते हैं। परीक्षा में derangement recurrence में swap और non-swap cases अलग करें।
First choose the (3) fixed objects and derange the remaining (5) objects. In exams use choose fixed plus derange rest for exactly fixed points.
Step 2
Why this answer is correct
The correct answer is A. \(^{8}C_3D_5\). First choose the (3) fixed objects and derange the remaining (5) objects. In exams use choose fixed plus derange rest for exactly fixed points.
Step 3
Exam Tip
पहले fixed (3) objects चुनें और बाकी (5) objects derange करें। परीक्षा में exactly fixed points के लिए choose fixed plus derange rest लगाएं।
Choose the correct letters and derange the remaining letters into wrong envelopes. In exams solve exactly correct letters using the fixed-point formula.
Step 2
Why this answer is correct
The correct answer is A. \(^{7}C_2D_5\). Choose the correct letters and derange the remaining letters into wrong envelopes. In exams solve exactly correct letters using the fixed-point formula.
Step 3
Exam Tip
सही letters चुनें और बाकी letters को wrong envelopes में derange करें। परीक्षा में exactly correct letters को fixed-point formula से करें।
There are (n-k-1) position pairs for (A,B) and (2) choices for their order. In exams count positions first in fixed-gap problems.
Step 2
Why this answer is correct
The correct answer is A. (2(n-k-1)(n-2)!). There are (n-k-1) position pairs for (A,B) and (2) choices for their order. In exams count positions first in fixed-gap problems.
Step 3
Exam Tip
(A,B) के position pairs (n-k-1) हैं और order के (2) choices हैं। परीक्षा में fixed gap में positions पहले गिनें।
There are (11-4-1=6) position pairs and (A,B) can be ordered in (2) ways. In exams there is no need to separately choose the people between them.
Step 2
Why this answer is correct
The correct answer is A. \(2\cdot6\cdot9!\). There are (11-4-1=6) position pairs and (A,B) can be ordered in (2) ways. In exams there is no need to separately choose the people between them.
Step 3
Exam Tip
Position pairs (11-4-1=6) हैं और (A,B) का order (2) तरीकों से हो सकता है। परीक्षा में बीच वाले people को अलग चुनने की जरूरत नहीं होती।
After fixing (A), the position of (B) is fixed and the remaining (8) people are arranged. In exams fix one person in one-direction circular gap problems.
Step 2
Why this answer is correct
The correct answer is A. (8!). After fixing (A), the position of (B) is fixed and the remaining (8) people are arranged. In exams fix one person in one-direction circular gap problems.
Step 3
Exam Tip
(A) को fix करने पर (B) की position fixed हो जाती है और बाकी (8) people arrange होते हैं। परीक्षा में one-direction circular gap में one person fix करें।
Total circular arrangements are (8!) and the adjacent block gives \(2\cdot7!\) ways. In exams handle circular not-adjacent by complement.
Step 2
Why this answer is correct
The correct answer is A. \(8!-2\cdot7!\). Total circular arrangements are (8!) and the adjacent block gives \(2\cdot7!\) ways. In exams handle circular not-adjacent by complement.
Step 3
Exam Tip
Total circular arrangements (8!) हैं और adjacent block \(2\cdot7!\) ways देता है। परीक्षा में circular not-adjacent को complement से करें।
Seat the men circularly in (7!) ways and place the women in the gaps in (8!) ways. In exams do not add an extra factor (2) in circular alternation.
Step 2
Why this answer is correct
The correct answer is A. \(7!\cdot8!\). Seat the men circularly in (7!) ways and place the women in the gaps in (8!) ways. In exams do not add an extra factor (2) in circular alternation.
Step 3
Exam Tip
पहले men को circularly (7!) ways में बैठाएं और gaps में women को (8!) ways में रखें। परीक्षा में circular alternate में extra (2) factor न लगाएं।
Only rotations are duplicates, so the circular count is ((10-1)!). In exams divide by (2) only after reading the reflection condition.
Step 2
Why this answer is correct
The correct answer is A. (9!). Only rotations are duplicates, so the circular count is ((10-1)!). In exams divide by (2) only after reading the reflection condition.
Step 3
Exam Tip
केवल rotations duplicate हैं इसलिए circular count ((10-1)!) है। परीक्षा में reflection condition पढ़कर ही (2) से divide करें।
In a bracelet, both rotations and reflections are considered the same. In exams use (\frac{(n-1)!}{2}) for bracelets.
Step 2
Why this answer is correct
The correct answer is A. \(\frac{9!}{2}\). In a bracelet, both rotations and reflections are considered the same. In exams use (\frac{(n-1)!}{2}) for bracelets.
Step 3
Exam Tip
Bracelet में rotations और reflections दोनों same मानी जाती हैं। परीक्षा में bracelet के लिए (\frac{(n-1)!}{2}) use करें।
When the last digit is fixed as (0), there is no leading-zero issue and (5) places are filled from (9) non-zero digits. In exams separate the zero-last case.
Step 2
Why this answer is correct
The correct answer is A. \(^{9}P_5\). When the last digit is fixed as (0), there is no leading-zero issue and (5) places are filled from (9) non-zero digits. In exams separate the zero-last case.
Step 3
Exam Tip
Last digit (0) fix होने पर first place पर zero issue नहीं रहता और (9) non-zero digits से (5) places भरते हैं। परीक्षा में zero-last case अलग करें।
There are (4) non-zero even choices for the last digit and (8) remaining non-zero choices for the first digit. In exams handle first and last restrictions together.
Step 2
Why this answer is correct
The correct answer is A. \(4\cdot8\cdot{}^{8}P_4\). There are (4) non-zero even choices for the last digit and (8) remaining non-zero choices for the first digit. In exams handle first and last restrictions together.
Step 3
Exam Tip
Last digit के (4) non-zero even choices हैं और first digit के (8) non-zero choices बचते हैं। परीक्षा में first और last restrictions को साथ संभालें।
Choose the even positions and then multiply even and odd choices independently. In exams choose positions first in exactly digit-type problems.
Step 2
Why this answer is correct
The correct answer is A. \(^{6}C_3\cdot4^3\cdot4^3\). Choose the even positions and then multiply even and odd choices independently. In exams choose positions first in exactly digit-type problems.
Step 3
Exam Tip
Even positions चुनें और फिर even तथा odd choices independently multiply करें। परीक्षा में exactly digit type में positions first choose करें।
Every symbol appearing is an onto condition and missing symbols are removed. In exams solve at least once by inclusion-exclusion.
Step 2
Why this answer is correct
The correct answer is A. (\sum_{i=0}^{5}(-1)^i{}^{5}C_i(5-i)8). Every symbol appearing is an onto condition and missing symbols are removed. In exams solve at least once by inclusion-exclusion.
Step 3
Exam Tip
हर symbol का आना onto condition है और missing symbols हटते हैं। परीक्षा में at least once को inclusion-exclusion से करें।
First choose (4) symbols and then form onto strings on them. In exams use choose set plus onto count for exactly distinct symbols.
Step 2
Why this answer is correct
The correct answer is A. (^{6}C_4\sum_{i=0}^{4}(-1)^i{}^{4}C_i(4-i)9). First choose (4) symbols and then form onto strings on them. In exams use choose set plus onto count for exactly distinct symbols.
Step 3
Exam Tip
पहले (4) symbols चुनें और फिर उन पर onto strings बनाएं। परीक्षा में exactly distinct symbols में choose set plus onto count करें।
A. (r) positions से selected (s) symbols पर onto assignments/Onto assignments from (r) positions to selected (s) symbols
Step 1
Concept
The Stirling part partitions positions into non-empty groups and (s!) assigns groups to symbols. In exams treat exactly used symbols as onto mapping.
Step 2
Why this answer is correct
The correct answer is A. (r) positions से selected (s) symbols पर onto assignments / Onto assignments from (r) positions to selected (s) symbols. The Stirling part partitions positions into non-empty groups and (s!) assigns groups to symbols. In exams treat exactly used symbols as onto mapping.
Step 3
Exam Tip
Stirling part positions को non-empty groups में बांटता है और (s!) groups को symbols assign करता है। परीक्षा में exactly used symbols को onto mapping समझें।
The exponents sum to (10) and the coefficient comes from the multinomial form. In exams treat powers as group sizes.
Step 2
Why this answer is correct
The correct answer is A. \(\frac{10!}{2!3!1!4!}\). The exponents sum to (10) and the coefficient comes from the multinomial form. In exams treat powers as group sizes.
Step 3
Exam Tip
Exponents का sum (10) है और coefficient multinomial form से मिलता है। परीक्षा में powers को group sizes मानें।
The cases are: choose three (x)'s or choose one \(x^2\) and one (x). In exams add all cases that form the same power.
Step 2
Why this answer is correct
The correct answer is A. \(^{10}C_3+10\cdot9\). The cases are: choose three (x)'s or choose one \(x^2\) and one (x). In exams add all cases that form the same power.
Step 3
Exam Tip
Cases हैं: तीन (x) चुनें या एक \(x^2\) और एक (x) चुनें। परीक्षा में same power बनाने वाले all cases जोड़ें।
A. चार (x), या दो (x) और एक \(x^2\), या दो \(x^2\)/Four (x)'s, or two (x)'s and one \(x^2\), or two \(x^2\)'s
Step 1
Concept
All disjoint choices that form exponent (4) must be added. In exams make exponent partitions for polynomial coefficients.
Step 2
Why this answer is correct
The correct answer is A. चार (x), या दो (x) और एक \(x^2\), या दो \(x^2\) / Four (x)'s, or two (x)'s and one \(x^2\), or two \(x^2\)'s. All disjoint choices that form exponent (4) must be added. In exams make exponent partitions for polynomial coefficients.
Step 3
Exam Tip
Exponent (4) बनाने वाले सभी disjoint choices जोड़ने पड़ते हैं। परीक्षा में polynomial coefficient में exponent partitions बनाएं।
The cube roots of unity filter is useful for separating modulo (3) classes. In exams identify three-step coefficient sums with an advanced filter.
Step 2
Why this answer is correct
The correct answer is A. Roots of unity filter. The cube roots of unity filter is useful for separating modulo (3) classes. In exams identify three-step coefficient sums with an advanced filter.
Step 3
Exam Tip
Modulo (3) classes अलग करने के लिए cube roots of unity filter उपयोगी है। परीक्षा में three-step coefficient sums को advanced filter से पहचानें।
When the consecutive ratio is less than (1), the coefficients start decreasing. In exams the inequality reverses after the peak.
Step 2
Why this answer is correct
The correct answer is A. (n-r<r+1). When the consecutive ratio is less than (1), the coefficients start decreasing. In exams the inequality reverses after the peak.
Step 3
Exam Tip
Consecutive ratio (1) से कम होने पर coefficients घटने लगते हैं। परीक्षा में peak के बाद inequality उलटी हो जाती है।
Unequal equal-combination indices are complementary. In exams set the sum of lower indices equal to the upper index.
Step 2
Why this answer is correct
The correct answer is A. (2r+6=n). Unequal equal-combination indices are complementary. In exams set the sum of lower indices equal to the upper index.
Step 3
Exam Tip
Unequal equal-combination indices complementary होते हैं। परीक्षा में lower indices का sum upper index के बराबर करें।
\(\frac{{}^{n}C_r}{{}^{n}C_{r-1}}=\frac{n-r+1}{r}\), and cross multiplication gives the relation. In exams keep the ratio direction correct.
Step 2
Why this answer is correct
The correct answer is A. (3n-10r+3=0). \(\frac{{}^{n}C_r}{{}^{n}C_{r-1}}=\frac{n-r+1}{r}\), and cross multiplication gives the relation. In exams keep the ratio direction correct.
Step 3
Exam Tip
\(\frac{{}^{n}C_r}{{}^{n}C_{r-1}}=\frac{n-r+1}{r}\) होता है और cross multiplication से relation मिलता है। परीक्षा में ratio direction सही रखें।
The cases for at least (2) special are exactly (2) and exactly (3). In exams direct cases are clear for a small special group.
Step 2
Why this answer is correct
The correct answer is A. \(^{3}C_2{}^{9}C_3+{}^{3}C_3{}^{9}C_2\). The cases for at least (2) special are exactly (2) and exactly (3). In exams direct cases are clear for a small special group.
Step 3
Exam Tip
At least (2) special के cases exactly (2) और exactly (3) हैं। परीक्षा में small special group में direct cases साफ रहते हैं।
