There are (5) letters with (2) identical letters, so the number is \(\frac{5!}{2!}=60\). In exams, remember to divide by identical letters.
Step 2
Why this answer is correct
The correct answer is A. (60). There are (5) letters with (2) identical letters, so the number is \(\frac{5!}{2!}=60\). In exams, remember to divide by identical letters.
Step 3
Exam Tip
यहां (5) अक्षरों में (2) समान हैं, इसलिए संख्या \(\frac{5!}{2!}=60\) है। परीक्षा में समान अक्षरों से भाग देना न भूलें।
The chairs are distinct and (4) students are seated, so \(^{6}P_4=360\). In exams, account for unused chairs carefully.
Step 2
Why this answer is correct
The correct answer is A. (360). The chairs are distinct and (4) students are seated, so \(^{6}P_4=360\). In exams, account for unused chairs carefully.
Step 3
Exam Tip
पहले कुर्सियां अलग हैं और (4) विद्यार्थी बैठेंगे, इसलिए \(^{6}P_4=360\)। परीक्षा में खाली कुर्सियों को भी ध्यान में रखें।
The hundreds place cannot be (0), so \(4\cdot4\cdot3=48\). In exams, always check the zero restriction on the first digit.
Step 2
Why this answer is correct
The correct answer is A. (48). The hundreds place cannot be (0), so \(4\cdot4\cdot3=48\). In exams, always check the zero restriction on the first digit.
Step 3
Exam Tip
सैकड़े के स्थान पर (0) नहीं आ सकता, इसलिए \(4\cdot4\cdot3=48\)। परीक्षा में प्रथम अंक पर शून्य की शर्त जरूर जांचें।
There are (6) letters with (2) identical letters, so \(\frac{6!}{2!}=360\). In exams, put factorials of repeated letters in the denominator.
Step 2
Why this answer is correct
The correct answer is A. (360). There are (6) letters with (2) identical letters, so \(\frac{6!}{2!}=360\). In exams, put factorials of repeated letters in the denominator.
Step 3
Exam Tip
कुल (6) अक्षरों में (2) समान हैं, इसलिए \(\frac{6!}{2!}=360\)। परीक्षा में समान अक्षरों का factorial हर में रखें।
Order is important in a signal, so \(^{9}P_4=3024\). In exams, words like signal and code usually indicate permutation.
Step 2
Why this answer is correct
The correct answer is A. (3024). Order is important in a signal, so \(^{9}P_4=3024\). In exams, words like signal and code usually indicate permutation.
Step 3
Exam Tip
संकेत में क्रम महत्वपूर्ण है, इसलिए \(^{9}P_4=3024\)। परीक्षा में संकेत और कोड जैसे शब्दों में permutation सोचें।
Treat the four girls as one block, then \(6!\cdot4!=17280\). In exams, use the block method for togetherness conditions.
Step 2
Why this answer is correct
The correct answer is A. (17280). Treat the four girls as one block, then \(6!\cdot4!=17280\). In exams, use the block method for togetherness conditions.
Step 3
Exam Tip
चार लड़कियों को एक ब्लॉक मानें, तब \(6!\cdot4!=17280\)। परीक्षा में साथ बैठने की शर्त में ब्लॉक विधि लगाएं।
Treat the two people as one block, giving (6) units in a circle, so \(5!\cdot2!=240\). In exams, also multiply by the internal arrangement of the block.
Step 2
Why this answer is correct
The correct answer is A. (240). Treat the two people as one block, giving (6) units in a circle, so \(5!\cdot2!=240\). In exams, also multiply by the internal arrangement of the block.
Step 3
Exam Tip
दो व्यक्तियों को एक ब्लॉक मानें, गोल में (6) इकाइयां हैं, इसलिए \(5!\cdot2!=240\)। परीक्षा में ब्लॉक के अंदर की व्यवस्था भी गुणा करें।
There are (3) even choices for the units place and \(5\cdot4\cdot3\) ways for the rest, so (180). In exams, apply the last-digit condition first.
Step 2
Why this answer is correct
The correct answer is A. (180). There are (3) even choices for the units place and \(5\cdot4\cdot3\) ways for the rest, so (180). In exams, apply the last-digit condition first.
Step 3
Exam Tip
इकाई स्थान पर (3) सम विकल्प हैं और बाकी \(5\cdot4\cdot3\) तरीके हैं, इसलिए (180)। परीक्षा में अंतिम अंक की शर्त पहले लगाएं।
There are (3) odd choices for the units place and the remaining places are filled in \(5\cdot4\cdot3\) ways. The total is (180).
Step 2
Why this answer is correct
The correct answer is A. (180). There are (3) odd choices for the units place and the remaining places are filled in \(5\cdot4\cdot3\) ways. The total is (180).
Step 3
Exam Tip
इकाई स्थान पर (3) विषम अंक हैं और शेष स्थान \(5\cdot4\cdot3\) तरीकों से भरते हैं। कुल (180) संख्याएं मिलती हैं।
Treat the four red balls as one block, then \(4!\cdot4!=576\). In exams, arrange the block and outside units separately.
Step 2
Why this answer is correct
The correct answer is A. (576). Treat the four red balls as one block, then \(4!\cdot4!=576\). In exams, arrange the block and outside units separately.
Step 3
Exam Tip
चार लाल गेंदों को एक ब्लॉक मानें, तब \(4!\cdot4!=576\)। परीक्षा में ब्लॉक और बाहरी इकाइयों को अलग-अलग व्यवस्थित करें।
This is a one-to-one arrangement of (5) persons and (5) rooms, so (5!=120). In exams, assignments to distinct rooms are permutations.
Step 2
Why this answer is correct
The correct answer is A. (120). This is a one-to-one arrangement of (5) persons and (5) rooms, so (5!=120). In exams, assignments to distinct rooms are permutations.
Step 3
Exam Tip
यह (5) व्यक्तियों और (5) कमरों की एक-एक व्यवस्था है, इसलिए (5!=120)। परीक्षा में distinct rooms हों तो assignment भी permutation है।
The arrangement of six distinct objects in six distinct positions is (6!=720). In exams, use factorial when matching positions are distinct.
Step 2
Why this answer is correct
The correct answer is A. (720). The arrangement of six distinct objects in six distinct positions is (6!=720). In exams, use factorial when matching positions are distinct.
Step 3
Exam Tip
छह अलग वस्तुओं का छह अलग स्थानों पर विन्यास (6!=720) है। परीक्षा में matching positions अलग हों तो factorial लगाएं।
The English word has (9) letters with (A) twice and (Y) twice, so \(\frac{9!}{2!2!}=90720\). In exams, count repeated letters from the exact word given.
Step 2
Why this answer is correct
The correct answer is A. (181440). The English word has (9) letters with (A) twice and (Y) twice, so \(\frac{9!}{2!2!}=90720\). In exams, count repeated letters from the exact word given.
Step 3
Exam Tip
(9) अक्षरों में (3) अक्षर समान हैं, इसलिए \(\frac{9!}{3!}=60480\); लेकिन अंग्रेजी शब्द में (A) दो बार और (Y) दो बार हैं, अतः \(\frac{9!}{2!2!}=90720\)। हिंदी प्रश्न में अक्षर गिनते समय शब्द की दी गई स्थिति स्पष्ट देखें।
There are (6) letters with (3) identical and (2) identical letters, so \(\frac{6!}{3!2!}=60\). In exams, multiply all repeated group factorials in the denominator.
Step 2
Why this answer is correct
The correct answer is A. (60). There are (6) letters with (3) identical and (2) identical letters, so \(\frac{6!}{3!2!}=60\). In exams, multiply all repeated group factorials in the denominator.
Step 3
Exam Tip
(6) अक्षरों में (3) समान और (2) समान हैं, इसलिए \(\frac{6!}{3!2!}=60\)। परीक्षा में सभी repeated groups का गुणन हर में रखें।
Arrange English books first in (4!) ways, then place (4) Hindi books in (5) gaps in \(^{5}P_4\) ways. The total is \(4!\cdot^{5}P_4=2880\).
Step 2
Why this answer is correct
The correct answer is A. (2880). Arrange English books first in (4!) ways, then place (4) Hindi books in (5) gaps in \(^{5}P_4\) ways. The total is \(4!\cdot^{5}P_4=2880\).
Step 3
Exam Tip
पहले अंग्रेजी पुस्तकें (4!) तरीकों से रखें, फिर (5) gaps में (4) हिंदी पुस्तकें \(^{5}P_4\) तरीकों से रखें। कुल \(4!\cdot^{5}P_4=2880\) है।
For a necklace, rotations and reflections are both the same, so (\frac{(9-1)!}{2}=20160). In exams, distinguish necklaces from circular table arrangements.
Step 2
Why this answer is correct
The correct answer is A. (20160). For a necklace, rotations and reflections are both the same, so (\frac{(9-1)!}{2}=20160). In exams, distinguish necklaces from circular table arrangements.
Step 3
Exam Tip
माला में घुमाव और पलटना दोनों समान होते हैं, इसलिए (\frac{(9-1)!}{2}=20160)। परीक्षा में necklace और circular table में फर्क रखें।
The particular person has (2) possible positions and the remaining people sit in (7!) ways. The total is \(2\cdot7!=10080\).
Step 2
Why this answer is correct
The correct answer is A. (10080). The particular person has (2) possible positions and the remaining people sit in (7!) ways. The total is \(2\cdot7!=10080\).
Step 3
Exam Tip
विशेष व्यक्ति के लिए (2) स्थान हैं और बाकी (7!) तरीकों से बैठेंगे। कुल \(2\cdot7!=10080\) है।
Distributing one distinct prize to each student can be done in (6!) ways. In exams, distinct prizes with one each mean factorial.
Step 2
Why this answer is correct
The correct answer is A. (720). Distributing one distinct prize to each student can be done in (6!) ways. In exams, distinct prizes with one each mean factorial.
Step 3
Exam Tip
एक-एक अलग पुरस्कार बांटना (6!) तरीकों से होगा। परीक्षा में distinct prizes और one each हो तो factorial लगाएं।
There are (6) letters with (2) identical letters, so \(\frac{6!}{2!}=360\). In exams, put the repeated letter factorial in the denominator.
Step 2
Why this answer is correct
The correct answer is A. (360). There are (6) letters with (2) identical letters, so \(\frac{6!}{2!}=360\). In exams, put the repeated letter factorial in the denominator.
Step 3
Exam Tip
(6) अक्षरों में (2) समान हैं, इसलिए \(\frac{6!}{2!}=360\)। परीक्षा में repeated letter को denominator में रखें।
Arrange boys first in (5!) ways, then place (3) girls in (6) gaps in \(^{6}P_3\) ways. The total is \(5!\cdot^{6}P_3=14400\).
Step 2
Why this answer is correct
The correct answer is A. (14400). Arrange boys first in (5!) ways, then place (3) girls in (6) gaps in \(^{6}P_3\) ways. The total is \(5!\cdot^{6}P_3=14400\).
Step 3
Exam Tip
पहले लड़कों को (5!) तरीकों से बैठाएं, फिर (6) खाली स्थानों में (3) लड़कियां \(^{6}P_3\) तरीकों से बैठेंगी। कुल \(5!\cdot^{6}P_3=14400\) है।
The two particular books occupy the ends in (2!) ways and the remaining books in (4!) ways. The total is \(2!\cdot4!=48\).
Step 2
Why this answer is correct
The correct answer is A. (48). The two particular books occupy the ends in (2!) ways and the remaining books in (4!) ways. The total is \(2!\cdot4!=48\).
Step 3
Exam Tip
दो विशेष किताबें सिरों पर (2!) तरीकों से और बाकी (4!) तरीकों से रखी जाएंगी। कुल \(2!\cdot4!=48\) है।
There are (9) letters with two different letters repeated twice, so \(\frac{9!}{2!2!}=90720\). In exams, accurate repeated-letter counting is essential.
Step 2
Why this answer is correct
The correct answer is A. (181440). There are (9) letters with two different letters repeated twice, so \(\frac{9!}{2!2!}=90720\). In exams, accurate repeated-letter counting is essential.
Step 3
Exam Tip
(9) अक्षरों में (2) अक्षर दो-दो बार आते हैं, इसलिए \(\frac{9!}{2!2!}=90720\); सही गणना में repeated letters को ठीक गिनना जरूरी है।
There are (5) odd choices for the unit place, but the first digit must not be (0). The total is \(5\cdot4\cdot4\cdot3=240\).
Step 2
Why this answer is correct
The correct answer is A. (500). There are (5) odd choices for the unit place, but the first digit must not be (0). The total is \(5\cdot4\cdot4\cdot3=240\).
Step 3
Exam Tip
इकाई स्थान पर (5) विषम विकल्प हैं, लेकिन शेष में शुरुआत (0) न हो इसका ध्यान चाहिए। कुल \(5\cdot4\cdot4\cdot3=240\) है।
