(8) अलग-अलग प्लेटों में से (2) प्लेटों को ऊपरी और निचले रैक में रखने के तरीके कितने हैं?
How many ways are there to place (2) plates from (8) different plates on the upper and lower racks?
#permutations
#class11
#easy
A (56)
B (16)
C (28)
D (64)
Explanation opens after your attempt
Step 1
Concept
The upper and lower racks are distinct, so \({}^{8}P_{2}=56\). Order is counted for distinct positions.
Step 2
Why this answer is correct
The correct answer is A. (56). The upper and lower racks are distinct, so \({}^{8}P_{2}=56\). Order is counted for distinct positions.
Step 3
Exam Tip
ऊपरी और निचला रैक अलग हैं इसलिए \({}^{8}P_{2}=56\)। अलग स्थानों में order गिना जाता है।
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\({}^{6}P_{5}\) का मान क्या है?
What is the value of \({}^{6}P_{5}\)?
#permutations
#class11
#easy
A (360)
B (120)
C (720)
D (720)
Explanation opens after your attempt
Step 1
Concept
\({}^{6}P_{5}=6\times5\times4\times3\times2=720\). For five places, take five decreasing factors.
Step 2
Why this answer is correct
The correct answer is D. (720). \({}^{6}P_{5}=6\times5\times4\times3\times2=720\). For five places, take five decreasing factors.
Step 3
Exam Tip
\({}^{6}P_{5}=6\times5\times4\times3\times2=720\) है। पाँच स्थान हों तो पाँच घटते गुणनखंड लें।
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शब्द (CODE) के अक्षरों से ऐसे कितने क्रम बनेंगे जिनमें अंतिम अक्षर (E) हो?
How many arrangements of the letters of (CODE) are possible if the last letter is (E)?
#permutations
#class11
#easy
A (24)
B (12)
C (6)
D (4)
Explanation opens after your attempt
Step 1
Concept
The last position is fixed and the remaining (3) letters can be arranged in (3!=6) ways. Separate the fixed position first.
Step 2
Why this answer is correct
The correct answer is C. (6). The last position is fixed and the remaining (3) letters can be arranged in (3!=6) ways. Separate the fixed position first.
Step 3
Exam Tip
अंतिम स्थान निश्चित है और शेष (3) अक्षर (3!=6) तरीकों से लगेंगे। fixed position को पहले अलग करें।
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(14) प्रतिभागियों में से प्रथम और द्वितीय वक्ता चुनने के तरीके कितने हैं?
In how many ways can first and second speakers be chosen from (14) participants?
#permutations
#class11
#easy
A (28)
B (182)
C (91)
D (196)
Explanation opens after your attempt
Step 1
Concept
First and second speakers are ordered positions, so \({}^{14}P_{2}=182\). Speaking order uses permutation.
Step 2
Why this answer is correct
The correct answer is B. (182). First and second speakers are ordered positions, so \({}^{14}P_{2}=182\). Speaking order uses permutation.
Step 3
Exam Tip
पहला और दूसरा वक्ता अलग क्रम वाले स्थान हैं इसलिए \({}^{14}P_{2}=182\)। speaking order में permutation लगता है।
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(4) अलग-अलग दिशासूचक कार्डों को ऊपर से नीचे रखने के तरीके कितने हैं?
In how many ways can (4) different direction cards be placed from top to bottom?
#permutations
#class11
#easy
A (24)
B (16)
C (8)
D (12)
Explanation opens after your attempt
Step 1
Concept
The top-to-bottom arrangement of (4) distinct cards is (4!=24). Use factorial in linear order.
Step 2
Why this answer is correct
The correct answer is A. (24). The top-to-bottom arrangement of (4) distinct cards is (4!=24). Use factorial in linear order.
Step 3
Exam Tip
ऊपर से नीचे (4) अलग कार्डों की व्यवस्था (4!=24) है। linear order में factorial प्रयोग करें।
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अंकों (0,1,5,7) से बिना पुनरावृत्ति कितनी (3)-अंकीय संख्याएँ बन सकती हैं?
How many (3)-digit numbers can be formed from digits (0,1,5,7) without repetition?
#permutations
#class11
#easy
A (24)
B (18)
C (9)
D (18)
Explanation opens after your attempt
Step 1
Concept
There are (3) choices for the hundreds place, then (3) and (2) choices, total (18). Zero is not allowed in the first place.
Step 2
Why this answer is correct
The correct answer is D. (18). There are (3) choices for the hundreds place, then (3) and (2) choices, total (18). Zero is not allowed in the first place.
Step 3
Exam Tip
सैकड़ा स्थान पर (3) विकल्प हैं, फिर (3) और (2) विकल्प मिलते हैं, कुल (18)। पहले स्थान पर (0) की अनुमति नहीं है।
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(6) अलग-अलग स्टेशनों में से शुरू और समाप्ति स्टेशन चुनने के तरीके कितने हैं?
In how many ways can a starting station and ending station be chosen from (6) different stations?
#permutations
#class11
#easy
A (15)
B (36)
C (30)
D (12)
Explanation opens after your attempt
Step 1
Concept
Starting and ending are ordered, so \({}^{6}P_{2}=30\). In a route, changing start and end changes the case.
Step 2
Why this answer is correct
The correct answer is C. (30). Starting and ending are ordered, so \({}^{6}P_{2}=30\). In a route, changing start and end changes the case.
Step 3
Exam Tip
शुरू और समाप्ति क्रम वाले हैं इसलिए \({}^{6}P_{2}=30\)। route में starting और ending बदलने से मामला बदलता है।
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शब्द (BANANA) के अलग-अलग अक्षर-क्रम कितने होंगे?
How many distinct letter arrangements are possible for the word (BANANA)?
#permutations
#class11
#easy
A (720)
B (60)
C (120)
D (30)
Explanation opens after your attempt
Step 1
Concept
(A) occurs three times and (N) twice, so \(\frac{6!}{3!2!}=60\). Divide by factorials of repeated groups.
Step 2
Why this answer is correct
The correct answer is B. (60). (A) occurs three times and (N) twice, so \(\frac{6!}{3!2!}=60\). Divide by factorials of repeated groups.
Step 3
Exam Tip
(A) तीन बार और (N) दो बार है इसलिए \(\frac{6!}{3!2!}=60\)। repeated groups के factorial से भाग दें।
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(5) अलग-अलग लिफाफों पर (5) अलग-अलग टिकट चिपकाने के तरीके कितने हैं?
In how many ways can (5) different stamps be pasted on (5) different envelopes?
#permutations
#class11
#easy
A (120)
B (25)
C (60)
D (10)
Explanation opens after your attempt
Step 1
Concept
(5) different stamps can be pasted on (5) different envelopes in (5!=120) ways. Factorial is useful in one-to-one matching.
Step 2
Why this answer is correct
The correct answer is A. (120). (5) different stamps can be pasted on (5) different envelopes in (5!=120) ways. Factorial is useful in one-to-one matching.
Step 3
Exam Tip
(5) अलग टिकट (5) अलग लिफाफों पर (5!=120) तरीकों से लगेंगे। one-to-one matching में factorial उपयोगी है।
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\({}^{15}P_{0}\) का मान क्या है?
What is the value of \({}^{15}P_{0}\)?
#permutations
#class11
#easy
A (15)
B (0)
C (225)
D (1)
Explanation opens after your attempt
Step 1
Concept
\({}^{n}P_{0}=1\), so \({}^{15}P_{0}=1\). Choosing zero objects is counted in one way.
Step 2
Why this answer is correct
The correct answer is D. (1). \({}^{n}P_{0}=1\), so \({}^{15}P_{0}=1\). Choosing zero objects is counted in one way.
Step 3
Exam Tip
\({}^{n}P_{0}=1\) इसलिए \({}^{15}P_{0}=1\)। शून्य वस्तु चुनने का एक तरीका माना जाता है।
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अंकों (2,4,6,8,9) से बिना पुनरावृत्ति कितनी (3)-अंकीय विषम संख्याएँ बनेंगी?
How many (3)-digit odd numbers can be formed from digits (2,4,6,8,9) without repetition?
#permutations
#class11
#easy
A (24)
B (60)
C (12)
D (20)
Explanation opens after your attempt
Step 1
Concept
Only (9) can be in the units place, and the other two places can be filled in \(4\times3=12\) ways. For odd numbers, check the units place first.
Step 2
Why this answer is correct
The correct answer is C. (12). Only (9) can be in the units place, and the other two places can be filled in \(4\times3=12\) ways. For odd numbers, check the units place first.
Step 3
Exam Tip
इकाई स्थान पर केवल (9) आएगा और बाकी दो स्थान \(4\times3=12\) तरीकों से भरेंगे। odd number में इकाई स्थान पहले देखें।
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(8) बच्चों में से (5) बच्चों को एक बेंच पर क्रम से बैठाने के तरीके कितने हैं?
In how many ways can (5) children from (8) children be seated on a bench in order?
#permutations
#class11
#easy
A (336)
B (6720)
C (56)
D (120)
Explanation opens after your attempt
Step 1
Concept
The answer is \({}^{8}P_{5}=8\times7\times6\times5\times4=6720\). Left-right order on a bench matters.
Step 2
Why this answer is correct
The correct answer is B. (6720). The answer is \({}^{8}P_{5}=8\times7\times6\times5\times4=6720\). Left-right order on a bench matters.
Step 3
Exam Tip
उत्तर \({}^{8}P_{5}=8\times7\times6\times5\times4=6720\) है। बेंच पर बाएँ-दाएँ क्रम महत्त्वपूर्ण होता है।
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शब्द (DAD) के अलग-अलग arrangements कितने हैं?
How many distinct arrangements are there for the word (DAD)?
#permutations
#class11
#easy
A (3)
B (6)
C (2)
D (1)
Explanation opens after your attempt
Step 1
Concept
(D) occurs twice, so there are \(\frac{3!}{2!}=3\) arrangements. Do not count repeated letters as distinct.
Step 2
Why this answer is correct
The correct answer is A. (3). (D) occurs twice, so there are \(\frac{3!}{2!}=3\) arrangements. Do not count repeated letters as distinct.
Step 3
Exam Tip
(D) दो बार है इसलिए \(\frac{3!}{2!}=3\) arrangements हैं। repeated letters को अलग-अलग न गिनें।
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अक्षरों (A,E,I,O,U) में से (2) अक्षरों का क्रम बनाना हो तो कितने तरीके होंगे?
If an order of (2) letters is to be made from (A,E,I,O,U), how many ways are possible?
#permutations
#class11
#easy
A (10)
B (25)
C (5)
D (20)
Explanation opens after your attempt
Step 1
Concept
This is \({}^{5}P_{2}=5\times4=20\). In ordered letters, (AE) and (EA) are considered different.
Step 2
Why this answer is correct
The correct answer is D. (20). This is \({}^{5}P_{2}=5\times4=20\). In ordered letters, (AE) and (EA) are considered different.
Step 3
Exam Tip
यह \({}^{5}P_{2}=5\times4=20\) है। ordered letters में (AE) और (EA) अलग माने जाते हैं।
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(10) अलग-अलग बैजों में से (4) बैजों को क्रम में लगाने के तरीके कितने हैं?
How many ways are there to arrange (4) badges from (10) different badges in order?
#permutations
#class11
#easy
A (5040)
B (210)
C (5040)
D (10000)
Explanation opens after your attempt
Step 1
Concept
\({}^{10}P_{4}=10\times9\times8\times7=5040\). For (4) places, take (4) decreasing factors.
Step 2
Why this answer is correct
The correct answer is C. (5040). \({}^{10}P_{4}=10\times9\times8\times7=5040\). For (4) places, take (4) decreasing factors.
Step 3
Exam Tip
\({}^{10}P_{4}=10\times9\times8\times7=5040\) है। (4) स्थानों के लिए (4) घटते गुणनखंड लें।
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शब्द (SMART) के सभी अक्षरों से कितने अलग क्रम बनेंगे?
How many distinct arrangements can be formed using all letters of (SMART)?
#permutations
#class11
#easy
A (60)
B (120)
C (25)
D (240)
Explanation opens after your attempt
Step 1
Concept
The word (SMART) has (5) distinct letters, so (5!=120). The full arrangement of distinct letters is factorial.
Step 2
Why this answer is correct
The correct answer is B. (120). The word (SMART) has (5) distinct letters, so (5!=120). The full arrangement of distinct letters is factorial.
Step 3
Exam Tip
(SMART) में (5) अलग अक्षर हैं इसलिए (5!=120)। distinct letters की पूरी arrangement factorial होती है।
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\({}^{14}P_{1}\) का मान क्या है?
What is the value of \({}^{14}P_{1}\)?
#permutations
#class11
#easy
A (14)
B (1)
C (196)
D (28)
Explanation opens after your attempt
Step 1
Concept
Since \({}^{n}P_{1}=n\), \({}^{14}P_{1}=14\). For one position, count available objects directly.
Step 2
Why this answer is correct
The correct answer is A. (14). Since \({}^{n}P_{1}=n\), \({}^{14}P_{1}=14\). For one position, count available objects directly.
Step 3
Exam Tip
\({}^{n}P_{1}=n\) होता है इसलिए \({}^{14}P_{1}=14\)। एक स्थान के लिए सीधे उपलब्ध वस्तुएँ गिनें।
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(6) अलग-अलग फोल्डरों को एक कतार में रखना है, पर एक विशेष फोल्डर हमेशा अंतिम स्थान पर रहेगा। कितने तरीके होंगे?
(6) different folders are to be arranged in a row, but one special folder always remains in the last position. How many ways are possible?
#permutations
#class11
#easy
A (720)
B (36)
C (60)
D (120)
Explanation opens after your attempt
Step 1
Concept
The last position is fixed and the remaining (5) folders can be arranged in (5!=120) ways. Arrange the remaining items after fixing the item.
Step 2
Why this answer is correct
The correct answer is D. (120). The last position is fixed and the remaining (5) folders can be arranged in (5!=120) ways. Arrange the remaining items after fixing the item.
Step 3
Exam Tip
अंतिम स्थान निश्चित है और शेष (5) फोल्डर (5!=120) तरीकों से लगेंगे। fixed item के बाद बचे items व्यवस्थित करें।
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अंकों (1,5,9) से पुनरावृत्ति की अनुमति होने पर कितनी (3)-अंकीय संख्याएँ बनेंगी?
How many (3)-digit numbers can be formed from digits (1,5,9) if repetition is allowed?
#permutations
#class11
#easy
A (9)
B (18)
C (27)
D (6)
Explanation opens after your attempt
Step 1
Concept
Each place has (3) choices, so \(3^3=27\). When repetition is allowed, choices do not decrease.
Step 2
Why this answer is correct
The correct answer is C. (27). Each place has (3) choices, so \(3^3=27\). When repetition is allowed, choices do not decrease.
Step 3
Exam Tip
हर स्थान पर (3) विकल्प हैं इसलिए \(3^3=27\)। repetition allowed हो तो विकल्प कम नहीं होते।
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(13) अभ्यर्थियों में से अध्यक्ष और उपाध्यक्ष चुनने के तरीके कितने हैं?
In how many ways can a president and vice-president be chosen from (13) candidates?
#permutations
#class11
#easy
A (26)
B (156)
C (78)
D (169)
Explanation opens after your attempt
Step 1
Concept
The two posts are distinct, so \({}^{13}P_{2}=13\times12=156\). Do not use combination when posts are different.
Step 2
Why this answer is correct
The correct answer is B. (156). The two posts are distinct, so \({}^{13}P_{2}=13\times12=156\). Do not use combination when posts are different.
Step 3
Exam Tip
दो पद अलग हैं इसलिए \({}^{13}P_{2}=13\times12=156\)। पद अलग हों तो combination नहीं लगाएं।
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शब्द (GATE) के अक्षरों से ऐसे कितने क्रम बनेंगे जिनमें (G) पहला अक्षर हो?
How many arrangements of the letters of (GATE) are possible if (G) is the first letter?
#permutations
#class11
#easy
A (6)
B (12)
C (24)
D (4)
Explanation opens after your attempt
Step 1
Concept
The first position is fixed and the remaining (3) letters can be arranged in (3!=6) ways. Apply the fixed condition first.
Step 2
Why this answer is correct
The correct answer is A. (6). The first position is fixed and the remaining (3) letters can be arranged in (3!=6) ways. Apply the fixed condition first.
Step 3
Exam Tip
पहला स्थान निश्चित है और शेष (3) अक्षर (3!=6) तरीकों से लगेंगे। fixed condition को पहले लागू करें।
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(9) सवालों में से (3) सवालों को पहले, दूसरे और तीसरे क्रम में हल करने के तरीके कितने हैं?
In how many ways can (3) questions from (9) questions be solved in first, second and third order?
#permutations
#class11
#easy
A (84)
B (729)
C (27)
D (504)
Explanation opens after your attempt
Step 1
Concept
The solving order matters, so \({}^{9}P_{3}=9\times8\times7=504\). Use permutation in ordered selection.
Step 2
Why this answer is correct
The correct answer is D. (504). The solving order matters, so \({}^{9}P_{3}=9\times8\times7=504\). Use permutation in ordered selection.
Step 3
Exam Tip
हल करने का क्रम महत्त्वपूर्ण है इसलिए \({}^{9}P_{3}=9\times8\times7=504\)। ordered selection में permutation लगाएं।
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(5) अलग-अलग पतंगों को (5) बच्चों में एक-एक देकर बाँटने के तरीके कितने हैं?
In how many ways can (5) different kites be distributed one each among (5) children?
#permutations
#class11
#easy
A (25)
B (60)
C (120)
D (10)
Explanation opens after your attempt
Step 1
Concept
(5) different kites can be distributed among (5) children in (5!=120) ways. Factorial appears in one-to-one distribution of distinct objects.
Step 2
Why this answer is correct
The correct answer is C. (120). (5) different kites can be distributed among (5) children in (5!=120) ways. Factorial appears in one-to-one distribution of distinct objects.
Step 3
Exam Tip
(5) अलग पतंग (5) बच्चों में (5!=120) तरीकों से बाँटी जा सकती हैं। अलग वस्तुओं के एक-एक वितरण में factorial आता है।
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अंकों (0,2,4,6,8) से बिना पुनरावृत्ति कितनी (2)-अंकीय सम संख्याएँ बन सकती हैं?
How many (2)-digit even numbers can be formed from digits (0,2,4,6,8) without repetition?
#permutations
#class11
#easy
A (20)
B (16)
C (10)
D (12)
Explanation opens after your attempt
Step 1
Concept
If the unit digit is (0), there are (4) ways, and if it is a non-zero even digit, there are \(4\times3=12\) ways, total (16). Make cases for zero questions.
Step 2
Why this answer is correct
The correct answer is B. (16). If the unit digit is (0), there are (4) ways, and if it is a non-zero even digit, there are \(4\times3=12\) ways, total (16). Make cases for zero questions.
Step 3
Exam Tip
इकाई पर (0) हो तो (4) तरीके और इकाई पर गैर-शून्य सम अंक हो तो \(4\times3=12\) तरीके हैं, कुल (16)। शून्य वाले प्रश्न में cases बनाएं।
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\({}^{7}P_{7}\) का मान क्या होगा?
What will be the value of \({}^{7}P_{7}\)?
#permutations
#class11
#easy
A (5040)
B (49)
C (720)
D (7)
Explanation opens after your attempt
Step 1
Concept
When (r=n), \({}^{n}P_{n}=n!\), so \({}^{7}P_{7}=5040\). Use factorial when all are selected.
Step 2
Why this answer is correct
The correct answer is A. (5040). When (r=n), \({}^{n}P_{n}=n!\), so \({}^{7}P_{7}=5040\). Use factorial when all are selected.
Step 3
Exam Tip
जब (r=n) हो तो \({}^{n}P_{n}=n!\), इसलिए \({}^{7}P_{7}=5040\)। पूरा चयन होने पर factorial लें।
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(6) अलग-अलग मेजों पर (2) विद्यार्थियों को बैठाने के तरीके कितने हैं?
In how many ways can (2) students be seated at (6) different desks?
#permutations
#class11
#easy
A (12)
B (15)
C (36)
D (30)
Explanation opens after your attempt
Step 1
Concept
The desks are distinct, so \({}^{6}P_{2}=6\times5=30\). Order is counted when seating at distinct places.
Step 2
Why this answer is correct
The correct answer is D. (30). The desks are distinct, so \({}^{6}P_{2}=6\times5=30\). Order is counted when seating at distinct places.
Step 3
Exam Tip
मेजें अलग हैं इसलिए \({}^{6}P_{2}=6\times5=30\)। अलग स्थानों पर बैठाने में क्रम गिना जाता है।
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शब्द (PAPER) में (P) दो बार आता है। इसके अलग-अलग अक्षर-क्रम कितने होंगे?
The word (PAPER) has (P) twice. How many distinct letter arrangements are possible?
#permutations
#class11
#easy
A (120)
B (30)
C (60)
D (20)
Explanation opens after your attempt
Step 1
Concept
Since (P) occurs twice, arrangements are \(\frac{5!}{2!}=60\). Divide by the factorial of the repeated letter.
Step 2
Why this answer is correct
The correct answer is C. (60). Since (P) occurs twice, arrangements are \(\frac{5!}{2!}=60\). Divide by the factorial of the repeated letter.
Step 3
Exam Tip
(P) दो बार समान है इसलिए व्यवस्थाएँ \(\frac{5!}{2!}=60\) होंगी। समान अक्षर के factorial से भाग दें।
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(7) अलग-अलग गीतों में से (3) गीतों को पहले, दूसरे और तीसरे स्थान पर बजाने के तरीके कितने हैं?
In how many ways can (3) songs from (7) different songs be played in first, second and third positions?
#permutations
#class11
#easy
A (35)
B (210)
C (21)
D (343)
Explanation opens after your attempt
Step 1
Concept
Playing order matters, so \({}^{7}P_{3}=7\times6\times5=210\). Changing the order changes the list.
Step 2
Why this answer is correct
The correct answer is B. (210). Playing order matters, so \({}^{7}P_{3}=7\times6\times5=210\). Changing the order changes the list.
Step 3
Exam Tip
बजाने का क्रम महत्त्वपूर्ण है इसलिए \({}^{7}P_{3}=7\times6\times5=210\)। क्रम बदलने से सूची बदल जाती है।
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\({}^{11}P_{2}\) का मान क्या है?
What is the value of \({}^{11}P_{2}\)?
#permutations
#class11
#easy
A (110)
B (121)
C (22)
D (55)
Explanation opens after your attempt
Step 1
Concept
\({}^{11}P_{2}=11\times10=110\). When (r=2), take two decreasing factors.
Step 2
Why this answer is correct
The correct answer is A. (110). \({}^{11}P_{2}=11\times10=110\). When (r=2), take two decreasing factors.
Step 3
Exam Tip
\({}^{11}P_{2}=11\times10=110\) होता है। (r=2) होने पर दो घटते गुणनखंड लें।
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अंकों (2,3,5,7) से बिना पुनरावृत्ति कितनी (3)-अंकीय संख्याएँ बनेंगी?
How many (3)-digit numbers can be formed from digits (2,3,5,7) without repetition?
#permutations
#class11
#easy
A (12)
B (16)
C (64)
D (24)
Explanation opens after your attempt
Step 1
Concept
For three places, there are \(4\times3\times2=24\) ways. Without repetition, each next choice decreases.
Step 2
Why this answer is correct
The correct answer is D. (24). For three places, there are \(4\times3\times2=24\) ways. Without repetition, each next choice decreases.
Step 3
Exam Tip
तीन स्थानों के लिए \(4\times3\times2=24\) तरीके हैं। बिना पुनरावृत्ति में हर अगला विकल्प घटता है।
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